Physics
Physics
1st Edition
Walker
ISBN: 9780133256925
Table of contents
Textbook solutions

All Solutions

Page 233: Lesson Check

Exercise 7
Step 1
1 of 2
$textbf{Solution: }$

Solve for the first case $m_{1} = m$:

In order to evaluate the magnitude of the Momentum, we use the formula that is given in the second yellow box:

$$
begin{align*}
p_{1} &= m_{1} ~ v \
&= m ~ v
end{align*}
$$

Solve for the second case $m_{2} =2 m$:

In order to evaluate the magnitude of the Momentum, we use the formula that is given in the second yellow box:

$$
begin{align*}
p_{2} &= m_{2} ~ v \
&=2 m ~ v \
&= 2 p_{1}
end{align*}
$$

So, in the case of doubling the mass of the object, the momentum will double, too.

Result
2 of 2
In the case of doubling the mass of the object, the momentum will double, too.
Exercise 8
Step 1
1 of 2
$textbf{Solution: }$

In order to evaluate the Momentum vector, we use the formula that is given in the first yellow box:

$$
begin{align*}
vec{p} &= vec{mv} \
&= m ~ vec{v} \
end{align*}
$$

So, the momentum vector is the product of the mass which is a scalar, and the velocity which is a vector. Therefore, the direction of the momentum vector is only depending on the direction of the velocity in the same direction.

Result
2 of 2
The momentum vector is the product of the mass which is a scalar, and the velocity which is a vector. Therefore, the direction of the momentum vector is only depending on the direction of the velocity in the same direction.
Exercise 9
Solution 1
Solution 2
Step 1
1 of 3
$textbf{Solution: }$

In order to evaluate the magnitude of the Momentum, we use the formula that is given in the second yellow box:

$$
begin{align*}
p &= m ~ v \
end{align*}
$$

So, in the case of doubling the mass of the object, the momentum will double, too.

Solve for the case that the speed of the object is $v_{1} = v$:

In order to evaluate the kinetic energy of the object, we use the following relation:

$$
begin{align*}
K.E_{1} &= dfrac{1}{2} ~ m ~ v_{1}^{2} \
&= dfrac{1}{2} ~ m ~ v^{2} \
end{align*}
$$

Step 2
2 of 3
$textbf{Solution: }$

Solve for the case that the speed of the object is $v_{2} = 2 v$:

By substituting by the double speed of the object, then we get

$$
begin{align*}
K.E_{2} &= dfrac{1}{2} ~ m ~ v^{2} \
&= dfrac{1}{2} ~ m ~ left( 2 v right)^{2} \
&= dfrac{1}{2} ~ m ~ 4 ~ v^{2} \
&= 4 ~ dfrac{1}{2} ~ m ~ v^{2} \
&= 4 K.E_{1}
end{align*}
$$

So, in the case of doubling the mass of the object, the kinetic energy will increase four times.

Result
3 of 3
In the case of doubling the speed of the object, the kinetic energy will increase four times.
Step 1
1 of 2
It is found that increasing the speed of object increases its momentum twice while Kinetic Energy increases four times. We know that $p=(mv)$ hence doubling the velocity will double the momentum $p=m(2v)$ but KE becomes.
$$
KE=frac{1}{2}m(2v)^2=4timesfrac{1}{2}mv^2
$$
Result
2 of 2
4 times
Exercise 10
Solution 1
Solution 2
Step 1
1 of 3
$textbf{Calculation: }$

In order to evaluate the Momentum vector, we use the formula that is given in the first yellow box:

$$
begin{align*}
vec{p} &= vec{mv} \
&= m ~ vec{v} \
end{align*}
$$

So, the total momentum vector of the two particles is equal to

$$
begin{align*}
vec{p_{T}} &= vec{p_{1}} + vec{p_{2}} \
&= m_{1} ~ vec{v_{1}} + m_{2} ~ vec{v_{2}} \
&= vec{0} \
end{align*}
$$

So, the zero momentum is a zero vector which means that either the sum magnitude of the two momentums is zero or the direction of the sum magnitude of the two momentums is zero.

In order to evaluate the kinetic energy, we use the following relation:

$$
begin{align*}
K.E &= dfrac{1}{2} ~ m ~ v^{2} \
end{align*}
$$

Step 2
2 of 3
$textbf{Solution: }$

So, the total kinetic energy of the two particles, which is a scalar, is equal to

$$
begin{align*}
K.E_{T} &= K.E_{1} + K.E_{2} \
&= dfrac{1}{2} ~ m_{1} ~ v_{1}^{2} + dfrac{1}{2} m_{2} ~ v_{2}^{2} \
end{align*}
$$

Since the zero momentum is a zero vector which means that either the summed magnitude of the two momentums is zero or the direction of the summed magnitude of the two momentums is zero. We have two cases, if the sum of the square of the two speeds is zero, then the total kinetic energy is zero. So, the counterexample, if the sum of the square of the two speeds isn’t zero and the two velocities are in opposite directions, then the total kinetic energy isn’t zero. Because the kinetic energy is a scalar (doesn’t depend on the direction).

Result
3 of 3
if the sum of the square of the two speeds isn’t zero and the two velocities are in opposite directions, then the total kinetic energy isn’t zero. Because the kinetic energy is a scalar (doesn’t depend on the direction).
Step 1
1 of 2
The sum of the momentums of the two objects with equal and opposite direction will be zero as they are vectors which will cancel each other. However the Kinetic energy will not be cancelled out because it is a scalar quantity so it implies that system with zero momentum may not have zero kinetic energy as well.
Result
2 of 2
Click here to see the explanation.
Exercise 11
Step 1
1 of 2
$textbf{Concept:}$
Product of mass and speed will give us momentum.

$textbf{Solution:}$ Multiply mass with speed to get $p$

$$
p=mv=(0.16,kg)(2.7,m/s)=0.43,kg.m/s
$$

$$
boxed{bf color{#4257b2}p=0.43,kg.m/s}
$$

Result
2 of 2
$$
p=0.43,kg.m/s
$$
Exercise 12
Step 1
1 of 3
$textbf{Given: }$

The mass of the school’s bus is $m_{s} = 1.82 times 10^{4} mathrm{~kg}$. The speed of the school’s bus is $v_{s} = 12.5 mathrm{~m/s}$. The mass of the baseball is $m_{b} = 0.142 mathrm{~kg}$.

$textbf{Required: }$

Finding the speed of the baseball, if the momentum of the school’s bus equals the momentum of the baseball.

Step 2
2 of 3
$textbf{Calculation: }$

Solve for the school’s bus:

In order to evaluate the magnitude of the Momentum of the school’s bus, we use the formula that is given in the second yellow box:

$$
begin{align*}
p_{s} &= m_{s} ~ v_{s} \
&= 1.82 times 10^{4} mathrm{~kg} times 12.5 mathrm{~m/s} \
&= 2.275 times 10^{5} mathrm{~kg cdot m/s} \
end{align*}
$$

Solve for the baseball:

In order to evaluate the magnitude of the Momentum of the baseball, we use the formula that is given in the second yellow box:

$$
begin{align*}
p_{b} &= m_{b} ~ v_{b} \
end{align*}
$$

Rearrange and solve for the speed of the baseball:

$$
begin{align*}
v_{b} &= dfrac{ p_{b} }{ m_{b} } \
&= dfrac{ 2.275 times 10^{5} mathrm{~kg cdot m/s} }{ 0.142 mathrm{~kg} } \
&= 1.602 times 10^{6} mathrm{~m/s}
end{align*}
$$

So, the speed of the baseball is $1.602 times 10^{6} mathrm{~m/s}$.

Result
3 of 3
The speed of the baseball is $1.602 times 10^{6} mathrm{~m/s}$.
Exercise 13
Step 1
1 of 3
$textbf{Given: }$

The speed of the running dog is $v = 1.9 mathrm{~m/s}$. The momentum of the running dog is $p = 17 mathrm{~kg cdot m/s}$.

$textbf{Required: }$

Finding the mass of the dog.

Step 2
2 of 3
$textbf{Calculation: }$

In order to evaluate the magnitude of the Momentum of the running dog, we use the formula that is given in the second yellow box:

$$
begin{align*}
p &= m ~ v \
end{align*}
$$

Rearrange and solve for the dog’s mass:

$$
begin{align*}
m &= dfrac{ p }{ v } \
&= dfrac{ 17 mathrm{~kg cdot m/s} }{1.9 mathrm{~m/s} } \
&= 8.947 mathrm{~kg}
end{align*}
$$

So, the dog’s mass is $8.947 mathrm{~kg}$.

Result
3 of 3
The dog’s mass is $8.947 mathrm{~kg}$.
Exercise 14
Step 1
1 of 3
$textbf{Given: }$

The mass of the cart $1$ is $m_{1} = 0.35 mathrm{~kg}$. The speed of the cart $1$ is $v_{1} = 1.2 mathrm{~m/s}$. The mass of the cart $2$ is $m_{2} = 0.61 mathrm{~kg}$. The speed of the cart $2$ is $v_{2} = 0.85 mathrm{~m/s}$.

$textbf{Required: }$

Finding the total momentum of the system.

Step 2
2 of 3
$textbf{Calculation: }$

In order to evaluate the Momentum vector, we use the formula that is given in the first yellow box:

$$
begin{align*}
vec{p} &= vec{mv} \
&= m ~ vec{v} \
end{align*}
$$

Let $hat{e}$ is the direction of cart $1$ motion. So, the total momentum vector of the two carts is equal to

$$
begin{align*}
vec{p_{T}} &= vec{p_{1}} + vec{p_{2}} \
&= m_{1} ~ vec{v_{1}} + m_{2} ~ vec{v_{2}} \
&= 0.35 mathrm{~kg} times 1.2 mathrm{~m/s} ~ hat{e} + 0.61 mathrm{~kg} times 0.85 mathrm{~m/s} ~ left( -hat{e} right) \
&= 0.35 mathrm{~kg} times 1.2 mathrm{~m/s} ~ hat{e} – 0.61 mathrm{~kg} times 0.85 mathrm{~m/s} ~hat{e} \
&= – left(0.0985 mathrm{~kg cdot m/s} right)~ hat{e}
end{align*}
$$

So, the total momentum of the system is $0.0985 mathrm{~kg cdot m/s}$ in the opposite direction of the cart $1$ motion.

Result
3 of 3
The total momentum of the system is $0.0985 mathrm{~kg cdot m/s} ~$ in the opposite direction of the cart $1$ motion.
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Chapter 1: Introduction to Physics
Section 1.1: Physics and the Scientific Method
Section 1.2: Physics and Society
Section 1.3: Units and Dimensions
Section 1.4: Basic Math for Physics
Page 38: Assessment
Page 41: Standardized Test Prep
Chapter 2: Introduction to Motion
Section 2.1: Describing Motion
Section 2.2: Speed and Velocity
Section 2.3: Position-Time Graphs
Section 2.4: Equation of Motion
Page 66: Assessment
Page 71: Standardized Test Prep
Page 45: Practice Problems
Page 47: Practice Problems
Page 47: Lesson Check
Page 49: Practice Problems
Page 52: Practice Problems
Page 53: Lesson Check
Page 56: Practice Problems
Page 57: Lesson Check
Page 59: Practice Problems
Page 60: Practice Problems
Page 62: Practice Problems
Page 62: Lesson Check
Chapter 3: Acceleration and Acceleration Motion
Section 3.1: Acceleration
Section 3.2: Motion with Constant Acceleration
Section 3.3: Position-Time Graphs for Constant Acceleration
Section 3.4: Free Fall
Page 105: Assessment
Page 111: Standardized Test Prep
Chapter 4: Motion in Two Dimensions
Section 4.1: Vectors in Physics
Section 4.2: Adding and Subtracting Vectors
Section 4.3: Relative Motion
Section 4.4: Projectile Motion
Page 144: Assessment
Page 149: Standardized Test Prep
Chapter 5: Newton’s Laws of Motion
Section 5.1: Newton’s Laws of Motion
Section 5.2: Applying Newton’s Laws
Section 5.3: Friction
Page 180: Assessment
Page 187: Standardized Test Prep
Chapter 6: Work and Energy
Section 6.1: Work
Section 6.2: Work and Energy
Section 6.3: Conservation of Energy
Section 6.4: Power
Page 220: Assessment
Page 227: Standardized Test Prep
Page 191: Practice Problems
Page 193: Practice Problems
Page 196: Lesson Check
Page 196: Practice Problems
Page 199: Practice Problems
Page 201: Practice Problems
Page 203: Practice Problems
Page 204: Practice Problems
Page 205: Practice Problems
Page 206: Lesson Check
Page 209: Practice Problems
Page 211: Lesson Check
Page 213: Practice Problems
Page 214: Practice Problems
Page 215: Practice Problems
Page 216: Lesson Check
Chapter 7: Linear Momentum and Collisions
Section 7.1: Momentum
Section 7.2: Impulse
Section 7.3: Conservation of Momentum
Section 7.4: Collisions
Page 260: Assessment
Page 265: Standardized Test Prep
Chapter 8: Rotational Motion and Equilibrium
Section 8.1: Describing Angular Motion
Section 8.2: Rolling Motion and the Moment of Inertia
Section 8.3: Torque
Section 8.4: Static Equilibrium
Page 300: Assessment
Page 305: Standardized Test Prep
Page 269: Practice Problems
Page 271: Practice Problems
Page 272: Practice Problems
Page 275: Practice Problems
Page 275: Lesson Check
Page 277: Practice Problems
Page 280: Lesson Check
Page 284: Practice Problems
Page 286: Practice Problems
Page 287: Practice Problems
Page 289: Lesson Check
Page 294: Practice Problems
Page 295: Practice Problems
Page 296: Lesson Check
Chapter 9: Gravity and Circular Motion
Section 9.1: Newton’s Law of Universal Gravity
Section 9.2: Applications of Gravity
Section 9.3: Circular Motion
Section 9.4: Planetary Motion and Orbits
Page 336: Assessment
Page 341: Standardized Test Prep
Chapter 10: Temperature and Heat
Section 10.1: Temperature, Energy, and Heat
Section 10.2: Thermal Expansion and Energy Transfer
Section 10.3: Heat Capacity
Section 10.4: Phase Changes and Latent Heat
Page 378: Assessment
Page 383: Standardized Test Prep
Chapter 11: Thermodynamics
Section 11.1: The First Law of Thermodynamics
Section 11.2: Thermal Processes
Section 11.3: The Second and Third Laws of Thermodynamics
Page 410: Assessment
Page 413: Standardized Test Prep
Chapter 12: Gases, Liquids, and Solids
Section 12.1: Gases
Section 12.2: Fluids at Rest
Section 12.3: Fluids in Motion
Section 12.4: Solids
Page 446: Assessment
Page 451: Standardized Test Prep
Chapter 13: Oscillations and Waves
Section 13.1: Oscillations and Periodic Motion
Section 13.2: The Pendulum
Section 13.3: Waves and Wave Properties
Section 13.4: Interacting Waves
Page 486: Assessment
Page 491: Standardized Test Prep
Chapter 14: Sound
Section 14.1: Sound Waves and Beats
Section 14.2: Standing Sound Waves
Section 14.3: The Doppler Effect
Section 14.4: Human Perception of Sound
Page 523: Assessment
Page 527: Standardized Test Prep
Page 495: Practice Problems
Page 496: Practice Problems
Page 500: Practice Problems
Page 501: Lesson Check
Page 503: Practice Problems
Page 504: Practice Problems
Page 506: Practice Problems
Page 506: Lesson Check
Page 510: Practice Problems
Page 511: Practice Problems
Page 512: Lesson Check
Page 514: Practice Problems
Page 516: Practice Problems
Page 517: Practice Problems
Page 519: Lesson Check
Chapter 15: The Properties of Lights
Section 15.1: The Nature of Light
Section 15.2: Color and the Electromagnetic Spectrum
Section 15.3: Polarization and Scattering of Light
Page 557: Assessment
Page 563: Standardized Test Prep
Chapter 16: Reflection and Mirrors
Section 16.1: The Reflection of Light
Section 16.2: Plane Mirrors
Section 16.3: Curved Mirrors
Page 590: Assessment
Page 595: Standardized Test Prep
Chapter 17: Refraction and Lenses
Section 17.1: Refraction
Section 17.2: Applications of Refraction
Section 17.3: Lenses
Section 17.4: Applications of Lenses
Page 629: Assessment
Page 635: Standardized Test Prep
Chapter 18: Interference and Diffraction
Section 18.1: Interference
Section 18.2: Interference in Thin Films
Section 18.3: Diffraction
Section 18.4: Diffraction Gratings
Page 668: Assessment
Page 673: Standardized Test Prep
Chapter 19: Electric Charges and Forces
Section 19.1: Electric Charge
Section 19.2: Electric Force
Section 19.3: Combining Electric Forces
Page 698: Assessment
Page 703: Standardized Test Prep
Chapter 20: Electric Fields and Electric Energy
Section 20.1: The Electric Field
Section 20.2: Electric Potential Energy and Electric Potential
Section 20.3: Capacitance and Energy Storage
Page 738: Assessment
Page 743: Standardized Test Prep
Chapter 21: Electric Current and Electric Circuits
Section 21.1: Electric Current, Resistance, and Semiconductors
Section 21.2: Electric Circuits
Section 21.3: Power and Energy in Electric Circuits
Page 775: Assessment
Page 781: Standardized Test Prep
Chapter 22: Magnetism and Magnetic Fields
Section 22.1: Magnets and Magnetic Fields
Section 22.2: Magnetism and Electric Currents
Section 22.3: The Magnetic Force
Page 810: Assessment
Page 815: Standardized Test Prep
Chapter 23: Electromagnetic Induction
Section 23.1: Electricity from Magnetism
Section 23.2: Electric Generators and Motors
Section 23.3: AC Circuits and Transformers
Page 844: Assessment
Page 849: Standardized Test Prep
Chapter 24: Quantum Physics
Section 24.1: Quantized Energy and Photons
Section 24.2: Wave-Particle Duality
Section 24.3: The Heisenberg Uncertainty Principle
Page 876: Assessment
Page 881: Standardized Test Prep
Chapter 26: Nuclear Physics
Section 26.1: The Nucleus
Section 26.2: Radioactivity
Section 26.3: Applications of Nuclear Physics
Section 26.4: Fundamental Forces and Elementary Particles
Page 944: Assessment
Page 947: Standardized Test Prep