WebAssign is not supported for this browser version. Some features or content might not work. System requirements

WebAssign

Welcome, demo@demo

(sign out)

Thursday, April 3, 2025 11:47 EDT

Home My Assignments Grades Communication Calendar My eBooks

My Cengage Home Notifications Help My Account

Physics - College, section 1, Fall 2019
My Assignments

Serway and Jewett - Physics for S&E 7/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 19 / 36

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

Last Saved : n/a Saving... ()

Print Assignment

Question

Points

1	2	3	4	5	6	7	8
–/1	–/2	–/5	–/12	–/2	–/9	–/3	–/2

Total

19/36 (52.8%)

Instructions

The WebAssign content for Physics for Scientists and Engineers 7/e by Serway and Jewett includes an extensive bank of more than 5,000 questions including end-of-chapter problems, interactive Active Figure questions, and tutorial problems offering feedback and hints to guide students to content mastery.

Instructors now have the option to display detailed solutions for specific problems after a specified number of submissions or after the assignment due date. These solutions (shown in blue below) use the randomized values from each student's version of selected problems.

WebAssign is the most widely used homework system in physics. Designed by physicists for physicists, WebAssign is easy to use, reliable, and a trusted companion to your teaching. Sign up for a Test Drive today!

Click here for a list of all of the questions coded in WebAssign for this textbook.

Question 1 is a traditional end-of-chapter problem with symbolic answer entry.

Question 2 is an end-of-chapter problem enhanced by adding a link to an Active Figure. The solution is displayed after the problem has been submitted

Question 3 shows an animated Active Figure question designed to combine traditional problem-solving skills with conceptual understanding.

Question 4 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 5 is a Quick Quiz question which serves as a checkpoint to help students test their understanding of physical concepts as they work through each chapter.

Question 6 is a Vector Problem. This is a standard end of chapter problem with a new vector identification part added to the beginning. This item is multiple choice with specific feedback for each choice.

Question 7 is an extra problem from the 6th edition, included in WebAssign.

Question 8 is a Math Review question from the math review appendix. This demo assignment allows many submissions and allows you to try another version of the same question for practice.

Assignment Submission

For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer.

Assignment Scoring

Your last submission is used for your score.

1. –/1 points SerPSE7 4.P.023. My Notes

A fireworks rocket explodes at height h, the peak of its vertical trajectory. It throws out burning fragments in all directions, but all at the same speed v. Pellets of solidified metal fall to the ground without air resistance. Find the smallest angle that the final velocity of an impacting fragment makes with the horizontal. (Answer using g, h, and v, as needed.)

Click here for help with symbolic formatting.

Viewing Saved Work Revert to Last Response

2. –/2 points was SerPSE7 4.P.061. My Notes

An enemy ship is on the east side of a mountain island, as shown in the figure. The enemy ship has maneuvered to within d₁ = 2137 m of the h = 1490 m high mountain peak and can shoot projectiles with an initial speed of v_i = 244 m/s. If the western shoreline is horizontally d₂ = 286 m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship?

less than m , or more than m

Refer to the Active Figure Simulation to review the concepts this question addresses.

Hint: Active Figure 4.7

Viewing Saved Work Revert to Last Response

3. –/5 points was SerPSE7 4.AF.10. My Notes

Active Figure 4.10 Projectile Launched From Level Ground

Instructions: The simulation below illustrates the motion of a projectile launched from level ground with an initial speed of 50 m/s. You can vary the initial angle with the blue slider. Click the fire button and observe the motion of the projectile. After the motion is complete, you can move your mouse to obtain the (x, y) coordinates of any point along the trajectory.

Display in a New Window | Explore with this Active Figure Explore
Consider the case of a projectile launched from level ground. Examples include: a golf ball hit from a tee on a level golf course, and a place-kicked football. Let us analyze the motion of the projectile to determine the following: (A) time to reach the maximum height, (B) the maximum height attained, (C) entire time of flight, (D) the range, and (E) the angle that gives the maximum range.

Conceptualize
Disregard any effect of forces exerted by the air, and consider the golf ball to be a point mass that starts with an initial velocity, upward and to the right. As it moves to the right with no force component acting on it in that direction, it simultaneously moves upward at first, but is pulled toward Earth by gravity throughout its flight. Use the simulation to display the resulting motion. Then later use the simulation to investigate each part of the solution.

Categorize
The initial velocity of the golf ball has components in both the x and y directions, so this is a problem about particle motion in two dimensions. We can model the golf ball as a particle with constant downward acceleration along the y direction combined with constant velocity in the x direction.

(A) Find the time to reach the maximum height:

Analyze The x-component and y-component of the motion during flight can be treated as independent of each other. At maximum height, the y-component of velocity is momentarily zero. So use the expression for motion in the y direction:

v_y = v_yi - gt

and set v_y to zero at the time t_{y max} of maximum height:

0 = v_i sinθ - gt_{y max}

Then solve for t_{y max}:

t_{y max}

v_i sinθ
black dot

Suppose the initial speed is 66.1 m/s and the launch angle is θ = 36°. In this case we find:

t_{y max} = s

(B) Find the maximum height y_max:

Analyze
The distance the particle moves upward while slowing with the acceleration -g is:

y_max = v_yit_{y max} - ½ g(t_{y max})²

Therefore, substitute in the expression for t_{y max} in part (a), and simplify, to obtain:

y_max

v_i² sin²θ
black dot

Suppose the initial speed is 66.1 m/s and the launch angle is θ = 36°. In this case, we have:

y max = m

(C) Find the entire time of flight:

Analyze
There are at least two approaches. One approach is to use the y-displacement equation, noting that the projectile starts from the ground and ends on the ground, where y is zero.

y = v_yi t - ½gt²

Hence, when the projectile has landed,

0 = v_i sinθ(t_flight) - ½g(t_flight)²

Cancel a factor of t from both terms, rearrange, and solve:

t_flight

2v_i sinθ

An easier way is to recognize the symmetry in this motion: the total time of flight is twice the time to reach the highest point. Check to see if this is so!

Suppose the initial speed is 66.1 m/s and the launch angle is θ = 36°. In this case, we have:

t_flight = s

(D) Find the range:

Analyze
While in flight, the projectile's horizontal velocity component is constant and x attains the maximum value when t = t_flight.

x = v_xi t + 0

Call this maximum value the range R:

R = v_i cosθ t_flight

Substituting the expression for t_flight found above and rearranging then gives:

2v_i²sinθcosθ
black dot

Suppose the initial speed is 66.1 m/s and the launch angle is θ = 36°. Find R.

R = m

(E) Find the angle that gives the maximum range:

Analyze
To find the maximum range, we wil use the trigonometric identity:

2 sinθ cosθ = sin(2θ)

Substituting this into the above expression for R gives:

v_i²sin(2θ)
black dot

The maximum R occurs when sin(2θ) is at its maximum, which occurs at:

θ_max = °

Use the simulation to verify this.

Finalize
While the y-component of the velocity is upward, it decreases by about 10 m/s during each second. So dividing the initial vertical velocity component (in m/s) by 10 gives an estimate of the time (in seconds) before the projectile momentarily stops to reverse direction.

Does that estimate agree roughly with your answer in part (a)?
Does twice that estimate agree roughly with your calculated time of flight, and does it look like it will give a horizontal velocity times the estimated time in agreement with your answer for the range?
Does the sign you obtain for the vertical velocity component at the end make sense?

Viewing Saved Work Revert to Last Response

4. –/12 points SerPSE7 4.AE.04. My Notes

A stone is thrown from the top of a building upward at an angle of 28.0° to the horizontal with an initial speed of 22.3 m/s as shown in the figure. The height of the building is 45.0 m.

(A) How long does it take the stone to reach the ground?

(B) What is the speed of the stone just before it strikes the ground?

A stone is thrown from the top of a building. Conceptualize Study the figure, in which we have indicated the trajectory and various parameters of the motion of the stone.

Categorize We categorize this problem as a projectile motion problem. The stone is modeled as a particle under constant acceleration in the y direction and a particle under constant velocity in the x direction.

Analyze We have the information x_i = y_i = 0, y_f = -45.0 m, a_y = -g, and v_i = 22.3 m/s (the numerical value of y_f is negative because we have chosen the top of the building as the origin).

(A) How long does it take the stone to reach the ground? Find the initial x and y components of the stone's velocity: v_xi = v_i cos θ_i = (22.3 m/s) cos 28.0° =

m/s
v_yi = v_i sin θ_i = (22.3 m/s) sin 28.0° = m/s Express the vertical position of the stone from the vertical component of Equation 4.9: y_f = y_i + v_yit + ½a_yt² Substitute numerical values: -45.0 m = (10.469 m/s)t + ½(-9.80 m/s²)t² Solve the quadratic equation for t: t = s

(B) What is the speed of the stone just before it strikes the ground? Use the y component of Equation 4.8 with t = 4.282 s to obtain the y component of the velocity of the stone just before it strikes the ground: v_yf = v_yi + a_yt Substitute numerical values: v_yf = 10.469 m/s + (-9.80 m/s²)(4.282 s) =

m/s Use this component with the horizontal component v_xf = v_xi = 19.690 m/s to find the speed of the stone at t = 4.282 s: v_f = √v_xf² + v_yf² = √(19.690 m/s)² + (-31.490 m/s)² = m/s

Finalize Is it reasonable that the y component of the final velocity is negative? Is it reasonable that the final speed is larger than the initial speed of 22.3 m/s?

The following questions present a twist on the scenario above to test your understanding.

Suppose another stone is thrown horizontally from the same building. If it strikes the ground 56 m away, find the following values.

(a) time of flight
s

(b) initial speed
m/s

(c) speed and angle with respect to the horizontal of the velocity vector at impact
m/s
°

If the stone were thrown harder, and left with 1.5 times the initial speed, you might expect it to go farther, but how exactly does that happen?

(d) Throwing the stone horizontally at 1.5 times the previous speed multiplies the time to reach the ground by what factor?

(e) The horizontal component of the velocity is multiplied by what factor?

(f) How many times farther does the stone land from the building?

Watch a Video Hint

Note from WebAssign
If a student makes one of several common mistakes on the Master It portion above, they will get feedback specific to the mistake they made. If their answer is incorrect but does not meet one of these specific conditions, they will still get generic numerical feedback as to how far off they are, or if they appear to have made just a sign or order of magnitude error, etc.

Try some of these values for part a to see what feedback you get!

Incorrect formula	Incorrect answer
	Answers may vary
	29.7
	883
	3.38
Correct formula	Correct answer
	3.03

Viewing Saved Work Revert to Last Response

5. –/2 points SerPSE7 4.QQ.04. My Notes

A particle moves in a circular path of radius r with speed v. It then increases its speed to 12v while traveling along the same circular path. By what factor has the centripetal acceleration of the particle changed?

By what factor has the period of the particle changed?

Viewing Saved Work Revert to Last Response

6. –/9 points SerPSE7 4.P.002. My Notes

A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by the following expressions.

x = (20.0 m/s)t
y = (5.50 m/s)t - (4.90 m/s²)t²

(a) Immediately after the golf ball is hit, in which of the following directions is the golf ball moving?

(b) Write a vector expression for the ball's position as a function of time, using the unit vectors i and j. (Use t, i and j as necessary.)
r = m

(c) By taking derivatives, obtain an expression for the velocity vector v as a function of time. (Use t, i and j as necessary.)
v = m/s

(d) By taking derivatives, obtain an expression for the acceleration vector a as a function of time.(Use t, i and j as necessary.)
a = m/s²

(e) Next use unit vector notation to write an expression for the position of the golf ball at t = 4.40 s.
r(4.40 s) = ( m ) i + ( m ) j

(f) Write an expression for the velocity at this time.
v(4.40 s) = ( m/s ) i + ( m/s ) j

(g) Write an expression for the acceleration at this time.
a(4.40 s) = ( m/s² ) j

Viewing Saved Work Revert to Last Response

7. –/3 points SerPSE7 4.XP.01. My Notes

At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of v_i = (3.00 i - 2.00 j) m/s and is at the origin. At t = 2.50 s, the particle's velocity is v = (6.60 i + 4.70 j) m/s.

(a) Find the acceleration of the particle at any time t. (Use t, i, and j as necessary.)
(

) m/s²

(b) Find its coordinates at any time t.
x = (

) i m
y = (

) j m

Viewing Saved Work Revert to Last Response

8. –/2 points SerPSE7 Demo.Math.Review. My Notes

Note from WebAssign: Each Serway textbook includes a Math Review Appendix. One such question is below. For an assignment of Math Review questions click here.

Quadratic Equations

Solve the following quadratic equation.

x² + 2x - 3 = 0

x = (smaller value)
x = (larger value)

Refer to the appendix

Appendix B

Viewing Saved Work Revert to Last Response

Home My Assignments

1	2
–/1	–/1
0/50	0/50

1	2	3	4	5
–/1	–/1	–/1	–/1	–/1
0/50	0/50	0/50	0/50	0/50

1	2	3	4	5	6	7	8	9	10	11	12
–/1	–/1	–/1	–/1	–/1	–/1	–/1	–/1	–/1	–/1	–/1	–/1
0/50	0/50	0/50	0/50	0/50	0/50	0/50	0/50	0/50	0/50	0/50	0/50

1	2
–/1	–/1
0/50	0/50