Tag Archives: College Physics Complete Solution Manual

Acceleration of a Revolving Ball – Uniform Circular Motion Example

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

Figure 1: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path.

Solution:

The linear velocity of the ball can be computed by dividing the total arc length traveled by the total time of travel. That is, the ball traveled 2 revolutions (twice the circumference of the circle) for 1 second. Thus,

\begin{align*}
\text{v} &= \frac{2\cdot2 \pi \text{r}}{\text{t}} \\
\\
& = \frac{4\pi \text{r}}{\text{t}} \\
\\
& = \frac{4\pi\left( 0.600\ \text{m} \right)}{1 \ \text{s}} \\
\\
& = 7.54 \ \text{m/s}
\end{align*}

Since the linear velocity has already been computed, we can now compute for the centripetal acceleration, ac.

\begin{align*}
\text{a}_\text{c} & = \frac{\text{v}^{2}}{\text{r}} \\
\\
& = \frac{\left( 7.54\ \text{m/s} \right)^{2}}{0.600\ \text{m}}\\
\\
& =94.8 \ \text{m/s}^{2}
\end{align*}

College Physics 2.51 – Time of the hiker to move out from a falling rock

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

Solution:

Part A

We know that the initial height, y_0 of the rock is 105 meters, and the initial velocity, v_0 is zero. We shall solve for the distance traveled by the rock for 1.5 seconds from the initial position first to find the height at detection.

The change in height is

\displaystyle \begin{aligned}
\Delta \text{y}&=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2 \\
&=\left( 0  \right)\left( 1.50 \ \text{s} \right)+\frac{1}{2}\left( 9.81\ \text{m/s}^{2} \right)\left( 1.50\ \text{s} \right)^{2}\\
&=0+11.036\ \text{m} \\
&=11.04 \ \text{m} 
\end{aligned}

So, the rock falls about 11.04 m from the initial height for 1.50 seconds. Therefore, the height of the rock above his head at this point is

\displaystyle \begin{aligned}
\text{y}&=\text{y}_{0}-\Delta \text{y} \\
&=105\ \text{m}-11.04\ \text{m} \\
&=93.96 \ \text{m}
\end{aligned}

Part B

We shall solve for the total time of travel, that is, from the initial position to his head. Then we shall subtract 1.50 s from that to solve for the unknown time of moving out. The total time of travel is

\begin{aligned}
\text{y} & =\frac{1}{2}\text{at}^{2} \\
&\text {Solving for t, we have}\\
\text{t}&=\sqrt{\frac{\text{2y}}{\text{a}}} \\
&=\sqrt{\frac{2\left( 105\ \text{m} \right)}{9.81 \ \text{m/s}^{2}}} \\
&=4.63 \ \text{s}

\end{aligned}

Therefore, to move out the hiker has about

\begin{aligned}
\text{t}&=4.63 \ \text{s}-1.50\ \text{s}\\
&=3.13\ \text{s}
\end{aligned}

College Physics 2.50 – Motion of a Jumping Kangaroo

A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

Part A

The motion of the kangaroo is under free-fall. We are looking for the initial velocity, and we know that the velocity in the highest position is zero.

From 

\begin{aligned}
\text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay},\\
\end{aligned}

we have

\begin{aligned}
\text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay}\\
\text{v}^2-2\text{ay} &= \left ( \text{v}_0\right)^2\\
\text{v}_0&=\sqrt{\text{v}^2-2\text{ay}}
\end{aligned}

Substituting the known values, 

\begin{aligned}
\text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \\
\text{v}_0&=\sqrt{0^2-2\left(-9.81 \text{m/s}^2\right)\left(2.50 \text{m}\right)}\\
\text{v}_0&= {\color{green}7.00 \  \text{m/s}}
\end{aligned}

Therefore, the vertical speed of the kangaroo when it leaves the ground is 7.00 m/s.

Part B

Since the motion of the kangaroo has uniform acceleration, we can use the formula

\text{y}=\text{v}_o\text{t}+\frac{1}{2}\text{a}\text{t}^2

The initial and final position of the kangaroo will be the same, so y is equal to zero. The initial velocity is 7.00 m/s, and the acceleration is -9.81 m/s2.

\begin{aligned}
\text{y} & =\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2\\
0 & = \left( 7.00\ \text{m/s} \right)\text{t}+\frac{1}{2}\left( -9.81\ \text{m/s}^{2} \right)\text{t}^2\\
0 & =7\text{t}-4.905\text{t}^{2}\\
7\text{t}-4.905\text{t}^{2}&=0 \\
\text{t}\left( 7-4.905\text{t} \right) & =0 \\
\text{t}=0 \qquad &\text{or} \qquad 7-4.905\text{t}=0 \\

\end{aligned}

Discard the time 0 since this refers to the beginning of motion. Therefore, we have

\begin{aligned}
7-4.905\text{t} &=0 \\
4.905\text{t} & = 7 \\
\text{t} & =\frac{7}{4.905} \\
 \text{t}&={\color{green}1.43 \  \text{s}} 
\end{aligned}

The kangaroo is about 1.43 seconds long in the air.

Solution Guides to College Physics by Openstax Chapter 13 Banner

Chapter 13: Temperature, Kinetic Theory, and the Gas Laws

Temperature

Problem 1

Problem 2

Problem 3

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Problem 5

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Thermal Expansion of Solids and Liquids

Problem 9

Problem 10

Problem 11

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Problem 13

Problem 14

Problem 15

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The Ideal Gas Law

Problem 22

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Problem 27

Problem 28

Problem 29

Problem 30

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Problem 35

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Problem 37

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Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

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Problem 48

Humidity, Evaporation, and Boiling

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

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Solution Guides to College Physics by Openstax Chapter 12 Banner

Chapter 12: Fluid Dynamics and Its Biological and Medical Applications

Flow Rate and Its Relation to Velocity

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

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Problem 7

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Bernoulli’s Equation

Problem 17

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The Most General Applications of Bernoulli’s Equation

Problem 25

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Viscosity and Laminar Flow; Poiseuille’s Law

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

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Problem 37

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Problem 39

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Problem 48

Problem 49

Problem 50

The Onset of Turbulence

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

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Problem 61

Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Solution Guides to College Physics by Openstax Chapter 11 Banner

Chapter 11: Fluid Statics

Density

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Pressure

Problem 11

Problem 12

Problem 13

Variation of Pressure with Depth in a Fluid

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

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Problem 23

Pascal’s Principle

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Gauge Pressure, Absolute Pressure, and Pressure Measurement

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Archimedes’ Principle

Problem 36

Problem 37

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Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

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Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

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Problem 53

Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action

Problem 54

Problem 55

Problem 56

Problem 57

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Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

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Problem 67

Pressures in the Body

Problem 68

Problem 69

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Problem 72

Problem 73

Problem 74

Problem 75

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Problem 79

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Problem 81

Problem 82

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Problem 87

Solution Guides to College Physics by Openstax Chapter 10 Banner

Chapter 10: Rotational Motion and Angular Momentum

Angular Acceleration

Problem 1

Problem 2

Problem 3

Problem 4

Kinematics of Rotational Motion

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Dynamics of Rotational Motion: Rotational Inertia

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Rotational Kinetic Energy: Work and Energy Revisited

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Angular Momentum and Its Conservation

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Collisions of Extended Bodies in Two Dimensions

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Gyroscopic Effects: Vector Aspects of Angular Momentum

Problem 48

Solution Guides to College Physics by Openstax Chapter 9 Banner

Chapter 9: Statics and Torque

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The Second Condition for Equilibrium

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

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Stability

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

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Applications of Statics, Including Problem-Solving Strategies

Problem 17

Problem 18

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Simple Machines

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

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Forces and Torques in Muscles and Joints

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

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Problem 37

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Problem 39

Problem 40

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Solution Guides to College Physics by Openstax Chapter 8 Banner

Chapter 8: Linear Momentum and Collisions

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Linear Momentum and Force

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

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Impulse

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

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Conservation of Momentum

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

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Elastic Collisions in One Dimension

Problem 28

Problem 29

Problem 30

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Inelastic Collisions in One Dimension

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

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Collisions of Point Masses in Two Dimensions

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

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Introduction to Rocket Propulsion

Problem 53

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Problem 64

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Solution Guides to College Physics by Openstax Chapter 7 Banner

Chapter 7: Work, Energy, and Energy Resources

Work: The Scientific Definition

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Kinetic Energy and the Work-Energy Theorem

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Gravitational Potential Energy

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Conservative Forces and Potential Energy

Problem 22

Problem 23

Nonconservative Forces

Problem 24

Problem 25

Conservation of Energy

Problem 26

Problem 27

Problem 28

Problem 29

Power

Problem 30

Problem 31

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Problem 33

Problem 34

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Work, Energy, and Power in Humans

Problem 44

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Problem 49

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Problem 51

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Problem 53

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World Energy Use

Problem 60

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Problem 70