HW3 Two-dimensional motion PDF

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HW3: Two-dimensional motion Due: 11:59pm on Sunday, February 14, 2021 You will receive no credit for items you complete after the assignment is due. Grading Policy

An Object Accelerating on a Ramp 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 x-y 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

Hint 1. Consider the change in velocity

when the object is at position 3?

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

Problem 3.03 A dragonfly flies from point

Part A

to point

along the path shown in the figure in 2.20

.

Find the

and

components of its position vector at point

.

Express your answers in meters to two significant figures separated by a comma. ANSWER: ,

= 2.0,1.0

,

Correct

Part B What is the magnitude of its position vector at

?

Express your answer in meters to two significant figures. ANSWER: = 2.20

Correct

Part C What is the direction of its position vector at

?

Express your answer in degrees to three significant figures. ANSWER: = 26.6

counterwise from

axis.

Correct

Part D Find the

and

components of the dragonfly's average velocity between

and

.

Express your answers in meter per second to two significant figures separated by a comma. ANSWER: ,

= 3.6,2.3

,

Correct

Part E What is the magnitude of its average velocity between these two points? Express your answers in meter per second to two significant figures. ANSWER: = 4.3

Correct

Part F What is the direction of its average velocity between these two points? Express your answers in degrees as an integer. ANSWER: = 32

counterwise from

axis.

Correct

Introduction to Projectile Motion

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 little different from the one-dimensional motions that you may already have studied. One of the most often-used techniques in physics is to divide two- and three-dimensional quantities into components. For instance, in projectile motion, a particle has some initial velocity . In general, this velocity can point in any direction in the xy plane and can have any magnitude. To make a problem more manageable, it is common to break up such a quantity into its x component and its y component . Consider a particle with initial velocity above the negative x axis.

that has magnitude 12.0

and is directed 60.0

Part A What is the x component

of ?

Express your answer in meters per second.

Hint 1. Determine which trignometric function to use Suppose that were directed at 60.0 above the positive x axis. Which of the following would give the correct relationship between (the magnitude of ) and (the magnitude of )? ANSWER:

Correct The fact that is actually directed at 60 above the negative x axis does not affect the magnitude of , but think about how it affects the sign of .

ANSWER: = -6.00

Correct

Part B What is the y component

of ?

Express your answer in meters per second.

Hint 1. Determine which trignometric function to use Which of the following gives the correct relationship between ) and (the magnitude of )? ANSWER:

Correct

(the magnitude of

ANSWER: = 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 Click on the image below to launch the video: Projectile Motion. Once you have watched the entire video, answer the graded follow-up questions. You can watch the video again at any point.

The motion diagram for a projectile is displayed, as are the motion diagrams for each component. The x component 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 constant-velocity motion. The vertical component exhibits constant-velocity 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 the case in which two balls are simultaneously dropped from a height of 5.0 . Click on the image below to launch the second video: Projectile Motion. This video shows the situation described in the problem. Once you have watched the entire video, answer the graded follow-up questions. You can watch the video again at any point.

Part D How long does it take for the balls to reach the ground? Use 10 magnitude of the acceleration due to gravity.

for the

Express your answer in seconds to one significant figure.

Hint 1. How to approach the problem The balls are released from rest at a height of at time . Using these numbers and the kinematic equation , you can determine the amount of time it takes for the balls to reach the ground.

ANSWER: = 1.0

Correct This situation, which you may 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 in which the horizontal component of the initial velocity is zero.

Part E Imagine that the ball on the left is given a nonzero initial velocity 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? 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 time , the ball will move horizontally 3.0 . You can assume that its initial x coordinate is .

ANSWER: = 3.0

Correct

Projectile Motion Tutorial Learning Goal: To understand how to apply the equations for one-dimensional motion to the x and y directions separately in order to derive standard formulas for the range and height of a projectile. A projectile is fired from ground level at time , at an angle with respect to the horizontal. It has an initial speed . In this problem we are assuming that the ground is level.

Part A Find the time Express gravity).

it takes the projectile to reach its maximum height in terms of

, , and

.

(the magnitude of the acceleration due to

Hint 1. A basic property of projectile motion The motion in the y direction is independent of the motion in the x direction because the force of gravity is acting in the direction, while no forces are acting in the direction (we are ignoring friction due to the air). Hint 2. What condition applies at the top of the trajectory? What is the value of the projectile's vertical velocity top of its trajectory? ANSWER: = 0

Hint 3. Vertical velocity as a function of time

when it has reached the

Find an expression for time.

, the projectile's vertical velocity, as a function of

Express your answer in terms of , gravity.

, , and , the acceleration due to

Hint 1. Determine the initial velocity in the y direction What is

, the initial y component of the velocity?

Provide your answer in terms of

and .

ANSWER: =

ANSWER: =

Hint 4. Putting it all together You now have a general expression for the vertical velocity as a function of time, and you also know the the vertical velocity at the top (i.e., at time ). This gives an equation that you can solve for . Hint 5. A list of possible answers There are four answers listed below, one of which is the correct answer for this question (the time for the projectile to reach the top of its trajectory). Once you have selected the correct answer, you will have the answer to Part A. (However, you will still need to enter this answer into the answer box ifor Part A.) ANSWER:

ANSWER: =

Correct

Part B Find , the time at which the projectile hits the ground after having traveled through a horizontal distance . Express the time in terms of

, , and .

Hint 1. Invoking the symmetry of the problem Because the projectile is fired over level ground, the time it takes for the projectile to reach its maximum height will be equal to the time it takes for the projectile to fall from its maximum height back to ground level. Thus, you can use your answer from Part A to quickly find the total flight time .

ANSWER:

=

Correct

Part C Find

, the maximum height attained by the projectile.

Express your answer in terms of

, , and .

Hint 1. Equation of motion Note that if you use the kinematic equation , then

and

Answer in terms of

. What is the initial vertical velocity

?

and .

ANSWER: =

ANSWER:

=

Correct

Part D Find the total distance (often called the range) traveled in the x direction; see the figure in the problem introduction.

Express the range in terms of

, , and .

Hint 1. When does the projectile hit the ground? The projectile reaches the ground at time

.

Hint 2. Determine the x position of the projectile as a function of time Give an expression for

, the x position of the particle as a function of time.

Express your answer in terms of ,

, and .

Hint 1. Acceleration in the x direction There is no acceleration in the x direction.

ANSWER: =

Hint 3. Finding the range Note that

.

Hint 4. A list of possible answers Choose the correct answer for from the following list. (You will still have to enter this answer into the main answer box for Part D.) ANSWER:

ANSWER:

=

Correct The actual formula for

is less important than how it is obtained:

1. Consider the motion in the x and y directions separately. 2. Find the time of flight from the motion in the y direction. 3. Find the x position at the end of the flight. This is the range. If you remember these steps, you can deal with many variants of the basic problem, such as a cannon on a hill that fires horizontally (i.e., the second half of the trajectory), a projectile that lands on a hill, or a projectile that must hit a moving target.

Projectile Motion--Conceptual 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 the base of the cliff, with positive upward.

at

ANSWER: =

Hint 2. Interpreting your equation Look at the formula you obtained for . At the projectile has been flying for 1/2 of the total flight time. According to your formula, what fraction of the way has it fallen down toward the ground?

ANSWER:

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.

± Delivering a Package by Air A relief airplane is delivering a food package to a group of people stranded on a very small island. The island is too small for the plane to land on, and the only way to deliver the package is by dropping it. The airplane flies horizontally with constant speed of 442 at an altitude of 625 . The positive x and y directions are defined in the figure. For all parts, assume that the "island" refers to the point at a distance from the point at which the package is released, as shown in the figure. Ignore the height of this point above sea level. Assume that the free-fall acceleration is = 9.80 .

Part A

After a package is dropped from the plane, how long will it take for it to reach sea level from the time it is dropped? Assume that the package, like the plane, has an initial velocity of 442 in the horizontal direction. Express your answer numerically in seconds. Neglect air resistance.

Hint 1. Knowns and unknowns: what are the initial conditions? Take the origin of the coordinate system to be at the point on the surface of the water directly below the point at which the package is released. The directions of the axes are shown in the figure in the problem introduction. In this coordinate system, what are the values of , , , of the package? Express your answers numerically and enter them, separated by commas, in the order , , , . Use units of meters and km/hour for distances and speeds, respectively.

Hint 1. Initial velocity in the y direction Because the package is dropped horizontally, the vertical component of its initial velocity is zero.

ANSWER: ,

,

,

= 0,625,442,0

Hint 2. What are the knowns and unknowns when the package hits the ground? Take the origin of the coordinate system to be at the point on the surface of the water directly below the point at which the package is released. The directions of the axes are shown in the figure in the problem introduction. Let be the time when the package hits the ground. In this coordinate system, 442 since there is no acceleration in the x direction. Which of the following values is/are known? Check all that apply. ANSWER:

Hint 3. Find the best equation to use Which of the equations below could you use to find the time hits the ground?

when the packet

Hint 1. How to determine which equation to use Only one of the quantities , , is known. Which one? You have to use the equation that contains this variable.

ANSWER:

ANSWER: = 11.3

Correc...