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Serway and Jewett - Physics for S&E 9/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 17 / 33

Due : Monday, January 28, 2030 00:00 EST

Last Saved : n/a Saving...  ()

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17/33 (51.5%)

  • Instructions

    Create your course assignments by selecting questions from our bank of end-of-section exercises, as well as enhanced interactive examples and tutorials.

    While doing their homework, students can link to the relevant interactive examples from the book and work through them again and again for additional practice before answering the question.

    Students will also find helpful links to online excerpts from their textbook and tutorials

    Read It - relevant textbook pages

    Master It - tutorials

    In this assignment we present several textbook question types found in Chapter 4 on Motion in Two Dimensions in Physics for Scientists and Engineers 9/e by Raymond A. Serway and John W. Jewett, Jr. published by Brooks/Cole.

    Question 1 is an Active Example which guides students through the process needed to master a concept. A new "Master It" question at the end provides a twist on the in-text Example to test student understanding.

    Question 2 is a PreLecture Exploration. It combines an Active Figure with conceptual and analytical questions that guide students to a deeper understanding and help promote a robust physical intuition.

    Question 3 is an Analysis Model Tutorial problem, which guides students through every step of the problem-solving process, driving students to see the important link between the situation in the problem and the mathematical representation of the situation.

    Question 4 is a Conceptual Question which serves as a concept check to help students test their understanding of physical concepts as they work through each chapter.

    Question 5 is an Objective Question.

    Questions 6, 7, and 8 are traditional end-of-chapter problems with symbolic answer entry.

    Question 9 is a Master It problem with a complete tutorial.

    Click here for a list of all of the questions coded in WebAssign. This demo assignment allows many submissions and allows you to try another version of the same question for practice wherever the problem has randomized values.

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1. 1/4 points  |  Previous Answers SerPSE9 4.AE.002. My Notes
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Example 4.2 The Long Jump
Mike Powell, current holder of the world long jump record of 8.95 m.
A long-jumper leaves the ground at an angle of 25.0° above the horizontal at a speed of 10.0 m/s.

(A) How far does he jump in the horizontal direction? (Assume his motion is equivalent to that of a particle.)

(B) What is the maximum height reached?
SOLVE IT
(A) How far does he jump in the horizontal direction?

Conceptualize The arms and legs of a long jumper move in a complicated way, but we will ignore this motion. We conceptualize the motion of the long jumper as equivalent to that of a simple projectile.

Categorize We categorize this example as a projectile motion problem. Because the initial speed and launch angle are given and because the final height is the same as the initial height, we further categorize this problem as satisfying the conditions for which Equations 4.12 and 4.13 can be used. This approach is the most direct way to analyze this problem, although the general methods that have been described will always give the correct answer.

Analyze
Use the horizontal range equation developed in this chapter to find the range of the jumper:
R =
vi2 sin 2θi
g
R = Correct: Your answer is correct. m
(B) What is the maximum height reached?

Analyze
We find the maximum height reached by using the equation:
h =
vi2 sin2θi
2g
h = Incorrect: Your answer is incorrect. m
Finalize Find the answers to parts (A) and (B) using the general method. The results should agree. Treating the long-jumper as a particle is an over-simplification. Nevertheless, the values obtained are consistent with experience in sports. We learn that we can model a complicated system such as a long-jumper as a particle and still obtain results that are reasonable.
MASTER IT HINTS: GETTING STARTED | I'M STUCK
A grasshopper jumps a horizontal distance of 2.00 from rest, with an initial velocity at an angle of 42.0° with respect to the horizontal.
(a) Find the initial speed of the grasshopper.
Incorrect: Your answer is incorrect. m/s Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.

(b) Find the maximum height reached by the grasshopper.
Incorrect: Your answer is incorrect. m Your response differs from the correct answer by more than 100%.
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2. 11/12 points  |  Previous Answers was SerPSE9 4.PLE.007. My Notes
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1/1 –/1 1/1 1/1 1/1 1/1 1/1 1/1 1/1 1/1 1/1 1/1
1/50 0/50 2/50 1/50 1/50 1/50 1/50 3/50 1/50 1/50 1/50 5/50
Total
11/12
 
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

PreLecture Exploration: Position and Time in Projectile Motion
This simulation shows a tennis ball launched from the origin by a tennis ball machine. The initial speed of the tennis ball is 30 m/s. Use the slider to set the launch angle or enter an angle value by clicking inside the degrees box and then click "fire" to set the tennis ball in motion. Click "pause" to stop the simulation at any point, and "clear" to remove traces of the projectile paths. Note that in this simulation, air resistance is ignored, although in reality air resistance is significant in the actual motion of a tennis ball.

Display in a New Window
Part 1 of 10 - Predictions
Several students are gathered around Kato, who is running the tennis ball machine. She wonders how high and how far the tennis ball can go. What launch angle do you predict Kato should choose to obtain the highest height possible? What angle should she choose so that the tennis ball travels the farthest? Enter your answers in the space below.
45 degrees
Key: Any thoughtful answer is acceptable.

Score: 1 out of 1

Comment:

Part 2 of 10 - Explore: Maximum Height
Kato runs the pitching machine several times, and all the students watch carefully to see what angle results in the maximum height. Run the simulation for several different angles to determine the angle of maximum height. What angle do you find?
    
Part 3 of 10 - Explore: Maximum Height
The students find that a 90° launch angle produces the maximum tennis ball height for a given launch speed. Kato asks the group why this is so. Choose the correct response.
     Correct: Your answer is correct.
Correct. As Chen recognized, the initial velocity is entirely in the vertical direction for a 90° launch, so the height of the tennis ball is greatest for a launch angle of 90°.
Part 4 of 10 - Explore: Maximum Height
Zuri challenges the group to use the equation for the time for any tennis ball to reach maximum height in the absence of air resistance, given by
ty,max = 
vi sin θ
g
,
 and the kinematic equation
yf = yi + vi(sin θi)t − 
1
2
gt2,
 to determine the maximum height, h, of the tennis ball, given the initial launch angle and speed. Choose the correct response.
     Correct: Your answer is correct.
Abel is correct. This equation can be used to determine the maximum height of a tennis ball in the absence of air resistance, given the initial velocity and launch angle. When Abel made his substitutions he got
h = vi (sin θi)t − 
1
2
g(ty,max)2
 which, using
ty,max = 
vi sin θi
g
 gave him
h = vi (sin θi)
vi sin θi
g
 − 
1
2
g
vi sin θi
g
2
 
.
 Simplifying this, he got
h = 
vi2(sin2θi)
g
 − 
vi2(sin2θi)
2g
,
 which reduced to this result.
Part 5 of 10 - Explore: Maximum Height
Darcy is not sure why, from
h = 
vi2 sin2θi
2g
,
 the height the tennis ball reaches is maximum when θ = 90°, and asks the group to explain. Choose the correct response.
     Correct: Your answer is correct.
Abel is correct. Because
sin 90° = 1,
 its maximum value, for any given initial speed, h is maximum.
Part 6 of 10 - Explore: Maximum Range
Although it is not possible to double the speed of the pitch machine in the simulation used here, Abel wonders how doubling the initial speed would affect the maximum height, and asks the group. Choose the correct response.
     Correct: Your answer is correct.
Kato is correct. She understood that there are two factors increasing maximum height—the greater initial speed (a factor of two) and the decreased time to reach maximum height (by a factor of ½, but in the denominator), which compound one another to make a factor of four altogether. Alternatively, she could have considered the kinematic equation for maximum height:
h = 
vi2 sin2θi
2g
.
 Because the initial speed vi is squared when the initial speed is doubled, this becomes
hnew = 
(2vi)2 sin2θi
2g
 or
4
vi2 sin2θi
2g
 = 4h.
Part 7 of 10 - Explore: Maximum Range
Kato again runs the pitching machine several times, and all the students watch carefully to see what angle results in the maximum horizontal range. Run the simulation for several different angles to determine the angle of maximum horizontal range. What angle do you find?
     Correct: Your answer is correct.
Correct. This is the angle for which the projectile travels farthest in the horizontal direction in the simulation.
Part 8 of 10 - Explore: Maximum Range
Chen reminds the group that in the absence of air resistance, the horizontal component of velocity of a projectile is constant, and challenges them to use the expression for the tennis ball to reach maximum height,
ty,max = 
vi sin θi
g
,
 and the constant horizontal velocity component, to find an equation for the range R of the tennis ball, in terms of the initial velocity and launch angle. They each come up with an answer. Choose the correct response.
     Correct: Your answer is correct.
Darcy is correct. She has derived the range equation for a projectile in the absence of air resistance on level ground.
Part 9 of 10 - Explore: Maximum Range
Kato asks why a 45° launch angle results in the maximum horizontal range. The four other students use
R = 
2vi2 sin θi cos θi
g
 to attempt an explanation. Choose the correct response.
     Correct: Your answer is correct.
Zuri is correct. He knows that for angles less than 45°, although the total time of flight of the projectile is longer, the horizontal component of velocity is less, so its horizontal displacement is less. For angles greater than 45°, although the horizontal component of velocity is greater, it requires less time to make the vertical component of the velocity zero, so the total time of flight of the projectile is shorter, so its horizontal displacement is also less than at a 45° launch angle.
Part 10 of 10 - Analyze
Use what you have learned to answer the following problem.

The initial speed of a tennis ball is 43.6 m/s and the launch angle is
θ = 36°.
 Neglect air resistance.
What is the maximum height, h, of the tennis ball?
33.5 Correct: Your answer is correct. seenKey

33.5

 m

What is the range, R, of the tennis ball?
184.481 Correct: Your answer is correct. seenKey

184

 m
Going Further
Kato observed that when she set the tennis ball machine first at 15° and then 75°, she noticed an interesting relationship. She also identified this same relationship with 30° and 60°. She challenged her friends to figure out the relationship, and they all came up with different answers. Choose the correct response.
     Correct: Your answer is correct.
Chen is correct. Complementary angles sum to 90°. From the range equation,
R = 
2vi2 sin θi cos θi
g
,
 the product
sin θ cos θ
 for angle θ is equal to
sin θ cos θ
 for its complement.)
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3. 1/1 points  |  Previous Answers SerPSE9 4.AMT.013. My Notes
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1/1 0/0 –/0 –/0 0/0 0/0 –/0 –/0 –/0 –/0 0/0 0/0 0/0 0/0 0/0 0/0 –/0 –/0 –/0 –/0 –/0
1/50 1/50 0/50 0/50 3/50 3/50 0/50 0/50 0/50 0/50 1/50 4/50 1/50 2/50 6/50 1/50 0/50 0/50 0/50 0/50 0/50
Total
1/1
 
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Analysis Model Tutorial
In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is h = 1.40 m. The mug slides off the counter and strikes the floor d = 1.10 m from the base of the counter.
(a) With what velocity did the mug leave the counter?

(b) What was the direction of the mug's velocity just before it hit the floor?
Part 1 of 10 - Conceptualize:
Once the mug leaves the counter, it becomes a projectile. We ignore air resistance because the mug is much denser than the surrounding air. The diagram shows an estimate of what the trajectory of the mug might look like. From the coordinate system in the diagram, we see that the origin is the point at which the mug leaves the counter. Based on our everyday experiences and the description of the problem, a reasonable speed of the mug would be a few m/s. It will hit the floor at an angle between
0°
 and
90°.
Part 2 of 10 - Categorize:
Now that we understand the problem and have made estimates of our expected results, we identify the analysis model(s) needed to solve the problem.

(1) Which of the following choices represents the best analysis model to use to describe the horizontal component of the motion of the beer mug?
     Correct: Your answer is correct.
Correct. Acceleration due to gravity is only in the −y-direction; the x (horizontal) velocity component is unchanged.
(2) Which of the following choices represents the best analysis model to use to describe the vertical component of the motion of the beer mug?
     Correct: Your answer is correct.
Correct. The acceleration due to gravity is in the −y-direction so the beer mug accelerates in the vertical direction.
Part 3 of 10 - Analyze:
(3) Based on the correct answer to question (1) above, what equation should we use to describe the horizontal motion of the mug?
    

(4) Based on the correct answer to question (2) above, what equation would be the most useful in describing the vertical motion of the mug and finding the time at which the mug strikes the floor?
    
Part 4 of 10 - Analyze: (cont.)
(5) Based on the correct answer to question (4) above, and identifying parameters in the equation that have a value of zero, what equation below correctly describes the time at which the mug strikes the floor?
     Correct: Your answer is correct.
Correct. This is the correct rearrangement of the equation that gives the time at which the mug lands on the floor.
Part 5 of 10 - Analyze: (cont.)
(6) Based on the correct answers to questions (3) and (5) above, and identifying parameters in the equation that have a value of zero, what equation below correctly describes the horizontal component of the velocity of the mug in terms of quantities given in the statement of the problem?
     Correct: Your answer is correct.
Correct. This is the correct combination of the equation from question (3) and the equation for the time from question (5).
Part 6 of 10 - Analyze: (cont.)
(a) With what velocity does the mug leave the counter?

(7) Based on the correct answer to question (6) above, substitute numerical values to find the speed with which the mug leaves the counter in the following.
vx = xf 
− 
g
2yf
 
 = 
(No Response) seenKey

1.10

 m
− 
(No Response) seenKey

9.8

 m/s2
2
(No Response) seenKey

-1.4

 m
 
 = (No Response) seenKey

2.06

 m/s
Part 7 of 10 - Analyze: (cont.)
(b) What was the direction of the mug's velocity just before it hit the floor?

(8) Based on the correct answer to question (2) above, and knowing what we now know, what equation could NOT be used to find the final velocity component of the mug in the vertical direction?
     Correct: Your answer is correct.
Correct. This equation does not involve the final velocity component in the vertical direction.
(9) Based on the correct answer to previous questions, and identifying parameters in the equation that have a value of zero, what equation below best describes the final vertical component of the velocity of the mug in terms of quantities given in the statement of the problem?
     Correct: Your answer is correct.
Correct. This equation is the result of carefully following the correct algebraic steps.
Part 8 of 10 - Analyze: (cont.)
(10) Based on the correct answer to question (9) above, substitute numerical values to find the final value of the vertical velocity component of the mug when it strikes the floor in the following.
vyf = 
−2gyf
 
 = 
−2
9.8 Correct: Your answer is correct. seenKey

9.8

 m/s2
-1.4 Correct: Your answer is correct. seenKey

-1.4

 m
 = -5.238 Correct: Your answer is correct. seenKey

-5.24

 m/s
Part 9 of 10 - Analyze: (cont.)
(11) Based on the correct answers to previous questions, what equation below correctly describes the angle that the final velocity vector makes with the horizontal?
     Correct: Your answer is correct.
Correct. This equation provides the correct value for the angle.
Part 10 of 10 - Analyze: (cont.)
(12) Based on the correct answer to previous questions above, substitute numerical values to find the value of the angle that the final velocity vector makes with respect to the horizontal when the mug strikes the floor in the following.
θ = tan−1
vyf
vx
 
  = tan−1
m/s
m/s
 
  =  °

This angle is measured as indicated in the figure. The mug's velocity is ° the horizontal when it strikes the floor.


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Which of the following particles, if any, have an acceleration?
Correct: Your answer is correct.


Explain your answer.

Score: 1 out of 1

Comment:

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A student throws a heavy red ball horizontally from a balcony of a tall building with an initial speed vi. At the same time, a second student drops a lighter blue ball from the balcony. Neglecting air resistance, which statement is true?
     Correct: Your answer is correct.
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The vector position of a particle varies in time according to the expression r with arrow = 6.20 î − 8.60t2 ĵ where r with arrow is in meters and t is in seconds.
(a) Find an expression for the velocity of the particle as a function of time. (Use any variable or symbol stated above as necessary.)
v with arrow =
F1,2​+β+cos(θ−90)
m/s

(b) Determine the acceleration of the particle as a function of time. (Use any variable or symbol stated above as necessary.)
a with arrow =
→y
m/s2

(c) Calculate the particle's position and velocity at t = 5.00 s.
r with arrow =
6.56→i−215→j
m
v with arrow =
ƒsƒasdƒ
m/s
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In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is h. The mug slides off the counter and strikes the floor at distance d from the base of the counter.
(a) With what velocity did the mug leave the counter? (Use any variable or symbol stated above along with the following as necessary: g.)
vxi =
​


(b) What was the direction of the mug's velocity just before it hit the floor? (Use any variable or symbol stated above as necessary.)
θ =
33​
Incorrect: Your answer is incorrect. below the horizontal
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A basketball star covers 2.75 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled precisely as that of a particle at his center of mass. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.95 m above the floor and is at elevation 0.950 m when he touches down again.
(a) Determine his time of flight (his "hang time").
Incorrect: Your answer is incorrect.
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. s

(b) Determine his horizontal velocity at the instant of takeoff.
Incorrect: Your answer is incorrect.
Your response differs from the correct answer by more than 100%. m/s

(c) Determine his vertical velocity at the instant of takeoff.
m/s

(d) Determine his takeoff angle.
Correct: Your answer is correct. ° above the horizontal

(e) For comparison, determine the hang time of a whitetail deer making a jump (see figure above) with center-of-mass elevations yi = 1.20 m, ymax = 2.55 m, and yf = 0.710 m.
s
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0/5
 
A home run is hit in such a way that the baseball just clears a wall 16 m high, located 116 m from home plate. The ball is hit at an angle of 31° to the horizontal, and air resistance is negligible. (Assume that the ball is hit at a height of 1.0 m above the ground.)
(a) Find the initial speed of the ball.
Incorrect: Your answer is incorrect.
Write the expressions for the horizontal and vertical positions as functions of time. Then use the equations to find the initial speed and the time when the baseball reaches the height of interest. m/s

(b) Find the time it takes the ball to reach the wall.
Incorrect: Your answer is incorrect. s

(c) Find the velocity components of the ball when it reaches the wall.
x-component Incorrect: Your answer is incorrect.
What is the initial velocity component in the horizontal direction? What force component changes the horizontal component of velocity during the flight of the baseball? m/s
y-component Incorrect: Your answer is incorrect.
Use the initial vertical component of velocity and the value of the acceleration due to gravity to calculate the final downward velocity component when the ball reaches the wall. m/s

Find the speed of the ball when it reaches the wall.
Incorrect: Your answer is incorrect.
Use the calculated values of the components of final velocity to calculate the final speed. m/s

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