Chapter 03 Homework

3/3/2014 Chapter 3 Homework

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014 1/21

Chapter 3 Homework

Due: 10:00pm on Friday, February 14, 2014

You will receive no credit for items you complete after the assignment is due. Grading Policy

Exercise 3.7

The coordinates of a bird flying in the xy-plane are given by and , where

and .

Part A

Calculate the velocity vector of the bird as a function of time.

Give your answer as a pair of components separated by a comma. For example, if you think the xcomponent is 3t and the y component is 4t, then you should enter 3t,4t. Express your answer using twosignificant figures for all coefficients.

ANSWER:

Correct

Part B

Calculate the acceleration vector of the bird as a function of time.

Give your answer as a pair of components separated by a comma. For example, if you think the xcomponent is 3t and the y component is 4t, then you should enter 3t,4t. Express your answer using twosignificant figures for all coefficients.

ANSWER:

Correct

Part C

Calculate the magnitude of the bird's velocity at .

Express your answer using two significant figures.

ANSWER:

Correct

40 C0 50 ND0

C NT

D NT

= 02 0 NT

= 0,-2.4 0 NT

0 T

= 5.4 2 NT

Typesetting math: 43%



Part D

Let the direction be the angle, that the vector makes with the +x-axis measured counterclockwise. Calculate thedirection of the bird's velocity at .

Express your answer in degrees using two significant figures.

ANSWER:

Correct

Part E

Calculate the magnitude of the bird's acceleration at .


ANSWER:

Correct

Part F

Calculate the direction of the bird's acceleration at .

ANSWER:

Correct

Part G

At , is the bird speeding up, slowing down or moving at constant speed?

ANSWER:

0 T

= -63 J

0 T

2.4 NT

0 T

= -90 J

0 T

speeding up

slowing down

moving at constant speed



Correct

Conceptual Problem about Projectile Motion

Learning Goal:

To understand projectile motion by considering horizontal constant velocity motion and vertical constant accelerationmotion independently.

Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near thesurface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance canbe neglected.

An object undergoing projectile motion near the surface of the earth obeys the following rules:

1. An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component ofits velocity, , is constant.

2. An object undergoing projectile motion moves vertically with a constant downward acceleration whosemagnitude, denoted by , is equal to 9.80 near the surface of the earth. Hence, the y component of

its velocity, , changes continuously.

3. An object undergoing projectile motion will undergo the horizontal and vertical motions described abovefrom the instant it is launched until the instant it strikes the ground again. Even though the horizontal andvertical motions can be treated independently, they are related by the fact that they occur for exactly thesame amount of time, namely the time the projectile is in the air.

The figure shows the trajectory (i.e., the path) of a ballundergoing projectile motion over level ground. The time

corresponds to the moment just after the ball islaunched from position and . Its launch

velocity, also called the initial velocity, is .

Two other points along the trajectory are indicated in thefigure.

One is the moment the ball reaches the peak of itstrajectory, at time with velocity . Its position at

this moment is denoted by or

since it is at its maximum height.

The other point, at time with velocity ,

corresponds to the moment just before the ball strikesthe ground on the way back down. At this time itsposition is , also known as ( since it is at its maximum horizontal range.

Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is waslaunched, as is the case here. Hence .

Part A

How do the speeds , , and (at times , , and ) compare?

ANSWER:

2

4

# NT

2

5

0

T0

N4

N5

2

0

2

4

5

4

5

NBY

0

2

4

5

4

NBY

5

N5

5

2

2

2

0

0

0



Correct

Here equals by symmetry and both exceed . This is because and include vertical speed as

well as the constant horizontal speed.

Consider a diagram of the ball at time . Recall that refersto the instant just after the ball has been launched, so it isstill at ground level ( ). However, it is already

moving with initial velocity , whose magnitude is and direction is

counterclockwise from the positive x direction.

Part B

What are the values of the intial velocity vector components and (both in ) as well as the acceleration

vector components and (both in )? Here the subscript 0 means "at time ."

Hint 1. Determining components of a vector that is aligned with an axis

If a vector points along a single axis direction, such as in the positive x direction, its x component will be itsfull magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. Ifthe vector points in the negative x direction, its x component will be the negative of its full magnitude.

Hint 2. Calculating the components of the initial velocity

Notice that the vector points up and to the right. Since "up" is the positive y axis direction and "to the

right" is the positive x axis direction, and will both be positive.

As shown in the figure, , , and are three sides of a right triangle, one angle of which is . Thus

and can be found using the definition of the sine and cosine functions given below. Recall that

= = > 0

= > = 0

= > > 0

> > > 0

> > = 0

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

0

0

N4

5

2

NT2

J EFHSFFT

2

4

2

5

NT

4

5

NT

0

2

2

4

2

5

2

4

2

5

2

J

2

4

2

5

NT

J EFHSFFT



and and note that

,

What are the values of and ?

Enter your answers numerically in meters per second separated by a comma.

ANSWER:

ANSWER:

Correct

Also notice that at time , just before the ball lands, its velocity components are (the same

as always) and (the same size but opposite sign from by symmetry). The

acceleration at time will have components (0, -9.80 ), exactly the same as at , as required by Rule

2.

The peak of the trajectory occurs at time . This is the point where the ball reaches its maximum height . At the

peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate.

Part C

What are the values of the velocity vector components and (both in ) as well as the acceleration

vector components and (both in )? Here the subscript 1 means that these are all at time .

ANSWER:

NT2

J EFHSFFT

TJOJ

MFOHUI PG PQQPTJUF TJEF

MFOHUI PG IZQPUFOVTF

2

5

2

DPTJ

MFOHUI PG BEKBDFOU TJEF

MFOHUI PG IZQPUFOVTF

2

4

2

2

4

2

5

15.0,26.0 NT

30.0, 0, 0, 0

0, 30.0, 0, 0

15.0, 26.0, 0, 0

30.0, 0, 0, -9.80

0, 30.0, 0, -9.80

15.0, 26.0, 0, -9.80

15.0, 26.0, 0, +9.80

0

NT2

4

NT2

5

2

5

0

NT

0

0

5

NBY

2

4

2

5

NT

4

5

NT

0



Correct

At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at thismoment it stops moving up and is about to move back down. This constitutes a downward-directed change invelocity, so the ball is accelerating downward even at the peak.

The flight time refers to the total amount of time the ball is in the air, from just after it is launched ( ) until just before itlands ( ). Hence the flight time can be calculated as , or just in this particular situation since .Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight timetraveling up to the peak and the other half traveling back down. The flight time is determined by the initial verticalcomponent of the velocity and by the acceleration. The flight time does not depend on whether the object is movinghorizontally while it is in the air.

Part D

If a second ball were dropped from rest from height , how long would it take to reach the ground? Ignore air

resistance.

Check all that apply.

Hint 1. Kicking a ball of cliff; a related problem

Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that theother is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the clifffirst? Ignore air resistance.

Hint 1. Comparing position, velocity, and acceleration of the two balls

Both balls start at the same height and have the same initial y velocity ( ) as well as the

same acceleration ( downward). They differ only in their x velocity (one is zero, the other

nonzero). This difference will affect their x motion but not their y motion.

ANSWER:

0, 0, 0, 0

0, 0, 0, -9.80

15.0, 0, 0, 0

15.0, 0, 0, -9.80

0, 26.0, 0, 0

0, 26.0, 0, -9.80

15.0, 26.0, 0, 0

15.0, 26.0, 0, -9.80

0

0

0

0

0

0

5

NBY

2

5

#



ANSWER:

Correct

In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as ittakes for it to fall from the peak back to the ground.

The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before itlands. Range is defined as x_2 - x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0.

Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity\texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can befound from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometricfunctions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also befound from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions.The following equations may be useful in solving projectile motion problems, but these equations apply only to aprojectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile.

flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}}

range: \large{R = v_x t_2 = \frac{v_0 2^ \sin(2\theta)}{g}}

In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launchangle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achievesthe maximum range for projectile motion over level ground is 45 degrees.

Part E

Which of the following changes would increase the range of the ball shown in the original figure?

Check all that apply.

ANSWER:

The ball that falls straight down strikes the ground first.

The ball that was kicked so it moves horizontally as it falls strikes the ground first.

Both balls strike the ground at the same time.

0

0

0

0

0

0

0

0



Correct

A solid understanding of the concepts of projectile motion will take you far, including giving you additionalinsight into the solution of projectile motion problems numerically. Even when the object does not land at thesame height from which is was launched, the rules given in the introduction will still be useful.Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be thebest choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The valueof the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be

appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets ormissiles. However, for problems that involve relatively dense projectiles moving close to the surface of theearth, these assumptions are reasonable.

Exercise 3.10

A daring 510-{\rm N} swimmer dives off a cliff with a running horizontal leap, as shown in the figure .

Part A

What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at thebottom, which is 1.75 {\rm m} wide and 9.00 {\rm m} below the top of the cliff?

ANSWER:

Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}.

Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}.

Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}.

Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}.

Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}.

v_0 = 1.29 {\rm m/s}



Correct

Exercise 3.12

A rookie quarterback throws a football with an initial upward velocity component of 15.9{\rm m/s} and a horizontalvelocity component of 19.9{\rm m/s} . Ignore air resistance.

Part A

How much time is required for the football to reach the highest point of the trajectory?

Express your answer using three significant figures.

ANSWER:

Correct

Part B

How high is this point?


ANSWER:

Correct

Part C

How much time (after it is thrown) is required for the football to return to its original level?


ANSWER:

Correct

Part D

How does this compare with the time calculated in part (a).

t_1 = 1.62 {\rm s}

h = 12.9 {\rm m}

t_2 = 3.24 {\rm s}




ANSWER:

Correct

Part E

How far has it traveled horizontally during this time?


ANSWER:

Correct

Exercise 3.13

A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so hedecides to try leaping it with his car. The side the car is on is 22.4{\rm m} above the river, while the opposite side is amere 1.6{\rm m} above the river. The river itself is a raging torrent 57.0{\rm m} wide.

Part A

How fast should the car be traveling just as it leaves the cliff in order just to clear the river and land safely on theopposite side?

ANSWER:

Correct

Part B

What is the speed of the car just before it lands safely on the other side?

ANSWER:

Correct

\large{\frac{t_2}{t_1}} = 2.00

x = 64.6 {\rm m}

v_0 = 27.7 {\rm m/s}

v = 34.3 {\rm m/s}



Exercise 3.15

Inside a starship at rest on the earth, a ball rolls off the top of a horizontal table and lands a distance D from the foot ofthe table. This starship now lands on the unexplored Planet X. The commander, Captain Curious, rolls the same ball offthe same table with the same initial speed as on earth and finds that it lands a distance 2.43 D from the foot of thetable.

Part A

What is the acceleration due to gravity on Planet X?

ANSWER:

Correct

Exercise 3.19

In a carnival booth, you win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a shelf above the pointwhere the quarter leaves your hand and is a horizontal distance of 2.1 {\rm m} from this point (the figure ). If you toss thecoin with a velocity of 6.4 {\rm{ m/s}} at an angle of 60 \^circabove the horizontal, the coin lands in the dish. You canignore air resistance.

Part A

What is the height of the shelf above the point where the quarter leaves your hand?


ANSWER:

g_{\rm X} = 1.66 {\rm m/s 2^}

H = 1.5 {\rm m}



Correct

Part B

What is the vertical component of the velocity of the quarter just before it lands in the dish?


ANSWER:

Correct

Problem 3.51

A jungle veterinarian with a blow-gun loaded with a tranquilizer dart and a sly 1.5-{\rm kg} monkey are each a height25{\rm m} above the ground in trees a distance 70{\rm m} apart. Just as the hunter shoots horizontally at the monkey,the monkey drops from the tree in a vain attempt to escape being hit.

Part A

What must the minimum muzzle velocity of the dart have been for the hunter to hit the monkey before it reached theground?


ANSWER:

Correct

Circular Launch

A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free fall, theball has a centripetal acceleration of magnitude 2 \texttip{g}{g}.

v_y = -0.89 {\rm m/s}

v = 31 {\rm m/s}



Part A

How far from the bottom of the chute does the ball land?

Your answer for the distance the ball travels from the end of the chute should contain \texttip{R}{R}.

Hint 1. Speed of ball upon leaving chute

How fast is the ball moving at the top of the chute?

Hint 1. Equation of motion

The centripetal acceleration for a particle moving in a circle is a_c = v^ 2/r, where \texttip{v}{v} is its

speed and \texttip{r}{r} is its instantaneous radius of rotation.

ANSWER:

Incorrect; Try Again; 5 attempts remaining

Hint 2. Time of free fall

How long is the ball in free fall before it hits the ground?

Express the free-fall time in terms of \texttip{R}{R} and \texttip{g}{g}.

Hint 1. Equation of motion

There is constant acceleration due to gravity, so you can use the general expression\large{y(t) = y_0 + v_{y0}t + \frac{a_y t 2^}{2}}.

\texttip{v}{v} = \sqrt{\left(2 g R\right)}



Write the values of \texttip{y_{\rm 0}}{y_0}, \texttip{v_{\rm y0}}{v_y0}, and \texttip{a_{\mit y}}{a_y}

(separated by commas) that are appropriate for this situation. Use the standard convention that \texttip{g}{g} is the magnitude of the acceleration due to gravity. Take y = 0 at the ground, and take

the positive y direction to be upward.

ANSWER:

Hint 2. Equation for the height of the ball

To find the time in free fall before the ball hits the ground, \texttip{t_{\rm f \hspace{1 pt}}}{t_f}, set the

general equation for the height equal to the height of the ground.

Answer in terms of \texttip{t_{\rm f \hspace{1 pt}}}{t_f}, \texttip{R}{R}, and \texttip{g}{g}.

ANSWER:

ANSWER:

Hint 3. Finding the horizontal distance

The horizontal distance follows from D = x(t_{\rm f \hspace{1 pt}}) = x_0 + v_{x0}t_{\rm f \hspace{1 pt}},

where x_0=0. \texttip{v_{\rm x0}}{v_x0} and \texttip{t_{\rm f \hspace{1 pt}}}{t_f} were found in Parts i and ii

respectively.

ANSWER:

Correct

Direction of Acceleration of Pendulum

Learning Goal:

To understand that the direction of acceleration is in the direction of the change of the velocity, which is unrelated to thedirection of the velocity.

The pendulum shown makes a full swing from -\pi/4 to + \pi/4. Ignore friction and assume that the string is massless.The eight labeled arrows represent directions to be referred to when answering the following questions.

\texttip{y_{\rm 0}}{y_0}, \texttip{v_{\rm y0}}{v_y0}, \texttip{a_{\mit y}}{a_y} = 2 R,0,-g

y(t_{\rm f \hspace{1 pt}})=0 = \large{2 R-\frac{g}{2} {t_{f}} {^2}}

\texttip{t_{\rm f \hspace{1 pt}}}{t_f} = \large{2 \sqrt{\frac{R}{g}}}

\texttip{D}{D} = \large{\sqrt{19.6 R} \sqrt{\frac{4R}{9.8}}}



Part A

Which of the following is a true statement about the acceleration of the pendulum bob, \texttip{\vec{a}}{a_vec}.

ANSWER:

Correct

Part B

What is the direction of \texttip{\vec{a}}{a_vec} when the pendulum is at position 1?

Enter the letter of the arrow parallel to \texttip{\vec{a}}{a_vec}.

Hint 1. Velocity at position 1

What is the velocity of the bob when it is exactly at position 1?

ANSWER:

Correct

Hint 2. Velocity of bob after it has descended

\texttip{\vec{a}}{a_vec} is equal to the acceleration due to gravity.

\texttip{\vec{a}}{a_vec} is equal to the instantaneous rate of change in velocity.

\texttip{\vec{a}}{a_vec} is perpendicular to the bob's trajectory.

\texttip{\vec{a}}{a_vec} is tangent to the bob's trajectory.

\texttip{v_{\rm 1}}{v_1} = 0 {\rm m/s}



What is the velocity of the bob just after it has descended from position 1?

ANSWER:

Correct

ANSWER:

Correct

Part C

What is the direction of \texttip{\vec{a}}{a_vec} at the moment the pendulum passes position 2?

Enter the letter of the arrow that best approximates the direction of \texttip{\vec{a}}{a_vec}.

Hint 1. Instantaneous motion

At position 2, the instantaneous motion of the pendulum can be approximated as uniform circular motion.What is the direction of acceleration for an object executing uniform circular motion?

ANSWER:

Correct

We know that for the object to be traveling in a circle, some component of its acceleration must be pointingradially inward.

Part D

What is the direction of \texttip{\vec{a}}{a_vec} when the pendulum reaches position 3?

Give the letter of the arrow that best approximates the direction of \texttip{\vec{a}}{a_vec}.

very small and having a direction best approximated by arrow D

very small and having a direction best approximated by arrow A

very small and having a direction best approximated by arrow H

The velocity cannot be determined without more information.

H

C



Hint 1. Velocity just before position 3

What is the velocity of the bob just before it reaches position 3?

ANSWER:

Hint 2. Velocity of bob at position 3

What is the velocity of the bob when it reaches position 3?

ANSWER:

ANSWER:

Correct

Part E

As the pendulum approaches or recedes from which position(s) is the acceleration vector \texttip{\vec{a}}{a_vec}

almost parallel to the velocity vector \texttip{\vec{v}}{v_vec}.

ANSWER:

Correct

Exercise 3.25

The earth has a radius of 6380 {\rm km} and turns around once on its axis in 24 {\rm h}.

very small and having a direction best approximated by arrow B

very small and having a direction best approximated by arrow C

very small and having a direction best approximated by arrow H

The velocity cannot be determined without more information.

\texttip{v_{\rm 3}}{v_3} = 0 {\rm m/s}

F

position 2 only

positions 1 and 2

positions 2 and 3

positions 1 and 3



Part A

What is the radial acceleration of an object at the earth's equator? Give your answer in {\rm{m/s}} 2^.

ANSWER:

Correct

Part B

What is the radial acceleration of an object at the earth's equator? Give your answer as a fraction of {\it g}.

ANSWER:

Correct

Part C

If a_{{\rm{rad}}} at the equator is greater than {\it g} , objects would fly off the earth's surface and into space. What

would the period of the earth's rotation have to be for this to occur?

ANSWER:

Correct

Exercise 3.29

A Ferris wheel with radius 14.0 {\rm m} is turning about a horizontal axis through its center (the figure ). The linear speedof a passenger on the rim is constant and equal to 6.40{\rm {\rm m/s}} .

a_{\rm rad} = 3.40102 {\rm{m/s}} 2^

a_{\rm rad} = 3.40103 {\it g}

T = 5070 {\rm s}



Part A

What is the magnitude of the passenger's acceleration as she passes through the lowest point in her circularmotion?

ANSWER:

Correct

Part B

What is the direction of the passenger's acceleration as she passes through the lowest point in her circular motion?

ANSWER:

Correct

Part C

What is the magnitude of the passenger's acceleration as she passes through the highest point in her circularmotion?

ANSWER:

a = 2.93 {\rm m/s 2^}

towards the center

outwards the center

a = 2.93 {\rm m/s 2^}



Correct

Part D

What is the direction of the passenger's acceleration as she passes through the highest point in her circularmotion?

ANSWER:

Correct

Part E

How much time does it take the Ferris wheel to make one revolution?

ANSWER:

Correct

Exercise 3.30

At its Ames Research Center, NASA uses its large 20-G centrifuge to test the effects of very large accelerations(hypergravity) on test pilots and astronauts. In this device, an arm 8.84 {\rm m} long rotates about one end in ahorizontal plane, and the astronaut is strapped in at the other end. Suppose that he is aligned along the arm with hishead at the outermost end. The maximum sustained acceleration to which humans are subjected in this machine istypically 12.5 {\it g}.

Part A

How fast must the astronaut's head be moving to experience this maximum acceleration?

ANSWER:

Correct

Part B

towards the center

outwards the center

T = 13.7 {\rm s}

v = 32.9 {\rm m/s}



What is the difference between the acceleration of his head and feet if the astronaut is 2.00 {\rm m} tall?

ANSWER:

Correct

Part C

How fast in rpm \left( {\rm rev/min} \right) is the arm turning to produce the maximum sustained acceleration?

ANSWER:

Correct

Problem 3.88

A projectile is thrown from a point P. It moves in such a way that its distance from P is always increasing.

Part A

Find the maximum angle above the horizontal with which the projectile could have been thrown. You can ignore airresistance.

ANSWER:

Correct

Score Summary:

Your score on this assignment is 98.3%.You received 13.77 out of a possible total of 14 points.

\Delta a = 27.7 {\rm m/s 2^}

\large{\frac{1}{T}} = 35.5 {\rm rpm}

\phi = 70.5 \^circ

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Chapter 03 Homework

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