Tag Archives: Momentum

College Physics by Openstax Chapter 8 Problem 8

A car moving at 10.0 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70.0 kg.

Solution:

Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes. In symbols, Newton’s second law of motion is defined to be

\text{F}_\text{net} = \frac{\Delta \textbf{p}}{\Delta t}

Moreover, Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity.

\textbf{p} = m \textbf{v}

So, the Newton’s second law becomes

\text{F}_\text{net} = \frac{m\Delta\textbf{v}}{\Delta t}

Substituting the given values, we have

\begin{align*} \text{F}_\text{net} = & \frac{m \Delta\textbf{v}}{\Delta t} \\ \text{F}_\text{net} = & \frac{\left( 70\ \text{kg} \right)\left( 10\ \text{m/s} \right)}{0.26\ \text{s}} \\ \text{F}_\text{net} = & 2692.3077 \ \text{N} \\ \text{F}_\text{net} = & 2.69 \times 10^{3}\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics Chapter 8 Problem 7

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600.0 m/s in a time of 2.00 ms (milliseconds)?

Solution:

Impulse, or change in momentum, equals the average net external force multiplied by the time this force acts:

\Delta \textbf{p}=F_{\text{net}} \Delta t

The unknown in this problem is the average force. We can solve the formula for force in terms of the other quantities.

F_{\text{net}} = \frac{\Delta \textbf{p}}{\Delta t}

We are given the following values:

\begin{align*} \Delta \textbf{p} & = m \Delta v = \left( 0.0300\ \text{kg} \right)\left( 600.0\ \text{m/s} \right) = 18.00\ \text{kg}\cdot \text{m/s} \\ \Delta t& = 2.00 \times 10^{-3}\ \text{s} \end{align*}

Substituting these given values, we can solve for the unknown.

\begin{align*} F_{\text{net}} & = \frac{\Delta \textbf{p}}{\Delta t} \\ F_{\text{net}} & = \frac{18.00\ \text{kg}\cdot \text{m/s}}{2.00 \times 10^{-3}\ \text{s}}\\ F_{\text{net}} & = 9000\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 8 Problem 6

The mass of Earth is $5.972 \times 10^{24}\ \text{kg}$ and its orbital radius is an average of $1.496 \times 10^{11}\ \text{m}$ . Calculate its linear momentum.

Solution:

Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum $\textbf{p}$ is defined to be

\textbf{p} = m \textbf{v}

where $m$ is the mass of the system and $\textbf{v}$ is its velocity.

For this problem, the mass has already been given to be $m=5.972 \times 10^{24}\ \text{kg}$ . We now proceed to calculating the velocity. We can solve the velocity of the using the formula

\textbf{v} = \frac{d}{t}

where $d$ is the total distance travelled, and $t$ is the total time of travel. The total distance traveled is just the circumference of the orbit, and the total time of travel is 1 year for 1 full revolution.

\begin{align*} \textbf{v} & = \frac{d}{t} \\ \textbf{v} & = \frac{2 \pi r}{t} \\ \textbf{v} & = \frac{2 \pi \left( 1.496\ \times 10^{11} \ \text{m} \right)}{365 \frac{1}{4} \ \text{days} \times \frac{24\ \text{hrs}}{\text{day}} \times \frac{3600\ \text{sec}}{1\ \text{hr}}} \\ \textbf{v} & = 29 785. 6783\ \text{m}/\text{s} \end{align*}

Therefore, its linear momentum is

\begin{align*} \textbf{p} & = m \textbf{v} \\ \textbf{p} & = \left( 5.972 \times 10^{24}\ \text{kg} \right) \left( 29 785. 6783\ \text{m}/\text{s} \right) \\ \textbf{p} & = 1.78 \times 10^{29}\ \text{kg} \cdot \text{m}/\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

College Physics by Openstax Chapter 8 Problem 5

A runaway train car that has a mass of 15,000 kg travels at a speed of 5.4 m/s down a track. Compute the time required for a force of 1500 N to bring the car to rest.

Solution:

\textbf{F}_{\text{net}} = \frac{\Delta \textbf{p}}{\Delta t} ,

where $\textbf{F}_{\text{net}}$ is the net external force, $\Delta \textbf{p}$ is the change in momentum, and $\Delta t$ is the change in time.

For this problem, we are given the following values:

\begin{align*} m & = 15000\ \text{kg} \\ \textbf{v}_{\text{initial}} & = 5.4\ \text{m}/\text{s} \\ \textbf{v}_{\text{final}} & = 0\ \text{m}/\text{s} \\ \textbf{F}_{\text{net}} & = 1500\ \text{N} \end{align*}

Substitute these given values in the equation above.

\begin{align*} \textbf{F}_{\text{net}} & = \frac{\Delta \textbf{p}}{\Delta t} \\ \Delta t & = \frac{\Delta \textbf{p}}{\textbf{F}_{\text{net}}} \\ \Delta t & = \frac{m \left( \Delta\textbf{v} \right)}{\textbf{F}_{\text{net}}} \\ \Delta t & = \frac{15000\ \text{kg} \left( 5.4\ \text{m}/\text{s} \right)}{1500\ \text{N}} \\ \Delta t & = 54\ s \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

It would take 54 seconds to stop the car.

College Physics by Openstax Chapter 8 Problem 4

(a) What is the momentum of a garbage truck that is 1.20×10⁴ kg and is moving at 30.0 m/s? (b) At what speed would an 8.00-kg trash can have the same momentum as the truck?

Solution:

Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum $\textbf{p}$ is defined as

\textbf{p}=m \textbf{v},

where $m$ is the mass of the system and $\textbf{v}$ is its velocity.

Part A. The momentum of the garbage truck

The garbage truck has the following quantities given:

\begin{align*} m_{\text{truck}} & = 1.20 \times 10^{4}\ \text{kg} \ \\ \textbf{v}_{\text{truck}} & = 30.0\ \text{m}/\text{s} \end{align*}

Substitute these values in the formula of momentum to solve for the momentum of the truck.

\begin{align*} \textbf{p} & = m \textbf{v} \\ \textbf{p} & = \left( 1.20 \times 10^{4}\ \text{kg} \right) \left( 30.0\ \text{m}/\text{s} \right) \\ \textbf{p} & = 360000\ \text{kg} \cdot \text{m}/\text{s} \\ \textbf{p} & = 3.60 \times 10^{5}\ \text{kg} \cdot \text{m}/\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. The Speed of a Trash to have the same Momentum as the Truck

The trash has the following properties:

\begin{align*} m & = 8.00\ \text{kg} \\ \textbf{p} & = 3.60 \times 10^{5}\ \text{kg} \cdot \text{m}/\text{s} \end{align*}

We now use the formula for momentum to solve for the unknown velocity.

\begin{align*} \textbf{p} & = m \textbf{v} \\ \textbf{v} & = \frac{\textbf{p}}{m} \\ \textbf{v} & = \frac{3.60 \times 10^{5}\ \text{kg} \cdot \text{m}/\text{s}}{8.00\ \text{kg}} \\ \textbf{v} & = 45000\ \text{m}/\text{s} \\ \textbf{v} & = 4.5 \times 10^{4}\ \text{m}/\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

College Physics by Openstax Chapter 8 Problem 3

(a) At what speed would a 2.00×10⁴-kg airplane have to fly to have a momentum of 1.60×10⁹ kg⋅m/s (the same as the ship’s momentum in the problem above)? (b) What is the plane’s momentum when it is taking off at a speed of 60.0 m/s? (c) If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship.

Solution:

Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum $\textbf{p}$ is defined as

\textbf{p}=m \textbf{v},

where $m$ is the mass of the system and $\textbf{v}$ is its velocity.

Part A. The Speed of an Airplane Given its Momentum

The airplane has the following quantities given:

\begin{align*} m_{\text{airplane}} & = 2.00 \times 10^{4}\ \text{kg} \\ \textbf{p}_{\text{airplane}} & = 1.60 \times 10^{9}\ \text{kg} \cdot \text{m}/\text{s} \end{align*}

Using the formula of momentum, we can solve for the velocity in terms of the other variables. We can then substitute the given quantities to solve for the velocity of the airplane.

\begin{align*} \textbf{p}_{\text{airplane}} & = m_{\text{airplane}} \textbf{v}_{\text{airplane}} \\ \textbf{v}_{\text{airplane}} & = \frac{\textbf{p}_{\text{airplane}}}{m_{\text{airplane}}} \\ \textbf{v}_{\text{airplane}} & = \frac{1.60 \times 10^{9}\ \text{kg} \cdot \text{m}/\text{s}}{2.00 \times 10^{4}\ \text{kg}} \\ \textbf{v}_{\text{airplane}} & = 80000\ \text{m}/\text{s} \\ \textbf{v}_{\text{airplane}} & = 8.00 \times 10^{4}\ \text{m}/\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. The Airplane’s Momentum When it is Taking Off

In this case, we are given the following properties of an airplane taking off:

\begin{align*} m_{\text{airplane}} & = 2.00 \times 10^{4}\ \text{kg} \\ \textbf{v}_{\text{airplane}} & = 60.0\ \text{m}/\text{s} \\ \end{align*}

The momentum of the airplane at this instance is calculated as

\begin{align*} \textbf{p}_{\text{airplane}} & = m_{\text{airplane}} \textbf{v}_{\text{airplane}} \\ \textbf{p}_{\text{airplane}} & = \left( 2.00 \times 10^{4}\ \text{kg} \right)\left( 60.0\ \text{m}/\text{s} \right) \\ \textbf{p}_{\text{airplane}} & = 1200000\ \text{kg} \cdot \text{m}/\text{s} \\ \textbf{p}_{\text{airplane}} & = 1.20 \times 10^{6}\ \text{kg} \cdot \text{m}/\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

Since the momentum of the airplane is 3 orders of magnitude smaller than the ship, the ship will not recoil very much. The recoil would be -0.01 m/s, which is probably not noticeable.

College Physics by Openstax Chapter 8 Problem 2

(a) What is the mass of a large ship that has a momentum of 1.60×10⁹ kg⋅m/s, when the ship is moving at a speed of 48.0 km/h? (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s.

Solution:

Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum $\textbf{p}$ is defined as

\textbf{p}=m \textbf{v},

where $m$ is the mass of the system and $\textbf{v}$ is its velocity.

Part A. The Mass of a Large Ship Given its Momentum

For this part, we are given the following quantities of the ship:

\begin{align*} \textbf{p}_{\text{ship}} & = 1.60 \times 10^{9}\ \text{kg} \cdot \text{m}/\text{s} \\ \textbf{v}_{\text{ship}} & = 48.0\ \text{km}/\text{h} = 13.3333\ \text{m}/\text{s} \end{align*}

From the formula of momentum, we can solve for $m$ in terms of $\textbf{p}$ and $\textbf{v}$ . Then we can substitute the given values to solve for the mass of the ship.

\begin{align*} \textbf{p}_{\text{ship}} & = m_{\text{ship}} \textbf{v}_{\text{ship}} \\ m_{\text{ship}} & = \frac{\textbf{p}_{\text{ship}}}{\textbf{v}_{\text{ship}}} \\ m_{\text{ship}} & = \frac{1.60 \times 10^{9}\ \text{kg} \cdot \text{m}/\text{s}}{13.3333\ \text{m}/\text{s}} \\ m_{\text{ship}} & = 120000000\ \text{kg} \\ m_{\text{ship}} & = 1.20 \times 10^{8} \ \text{kg}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. Comparing the Momentum of a Large Ship with an Artillery Shell

The artillery shell has a mass of 1100 kilograms and a speed of 1200 m/s. Therefore, its momentum is

\begin{align*} \textbf{p}_{\text{shell}} & = m_{\text{shell}} \textbf{v}_{\text{shell}} \\ \textbf{p}_{\text{shell}} & = \left( 1100\ \text{kg} \right)\left( 1200\ \text{m}/\text{s} \right) \\ \textbf{p}_{\text{shell}} & = 1320000\ \text{kg} \cdot \text{m}/\text{s} \\ \textbf{p}_{\text{shell}} & = 1.32 \times 10^{6}\ \text{kg} \cdot \text{m}/\text{s} \end{align*}

Comparing the momentum of the large ship and the shell, we have.

\begin{align*} \frac{\textbf{p}_{\text{ship}}}{\textbf{p}_{\text{shell}}} & = \frac{1.60 \times 10^{9}\ \text{kg} \cdot \text{m}/\text{s}}{1.32 \times 10^{6}\ \text{kg} \cdot \text{m}/\text{s}} \\ \frac{\textbf{p}_{\text{ship}}}{\textbf{p}_{\text{shell}}} & = 1212.1212 \\ \textbf{p}_{\text{ship}} & = 1212.1212\ \textbf{p}_{\text{shell}} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The ship has a momentum that is about 1200 times the momentum of the artillery shell.

College Physics by Openstax Chapter 8 Problem 1

(a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of 7.50 m/s. (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of 600 m/s. (c) What is the momentum of the 90.0-kg hunter running at 7.40 m/s after missing the elephant?

Solution:

Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum $\textbf{p}$ is defined as

\textbf{p}=m \textbf{v},

where $m$ is the mass of the system and $\textbf{v}$ is its velocity.

Part A. The Momentum of the Elephant

We are given the mass and the velocity of the elephant, so we can just directly substitute these values in the formula for momentum.

\begin{align*} \textbf{p}_{\text{elephant}} & = m_{\text{elephant}} \textbf{v}_{\text{elephant}} \\ \textbf{p}_{\text{elephant}} & = \left( 2000\ \text{kg} \right)\left( 7.50\ \text{m}/\text{s} \right) \\ \textbf{p}_{\text{elephant}} & = 15000\ \text{kg} \cdot \text{m}/\text{s} \\ \textbf{p}_{\text{elephant}} & = 1.50 \times 10 ^{4} \ \text{kg} \cdot \text{m}/\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. Comparing the momentum of the elephant in part A with the momentum of a tranquilizer

First, we need to calculate the momentum of the tranquilizer.

\begin{align*} \textbf{p}_{\text{tranquilizer}} & = m_{\text{tranquilizer}} \textbf{v}_{\text{tranquilizer}} \\ \textbf{p}_{\text{tranquilizer}} & = \left( 0.0400\ \text{kg} \right)\left( 600\ \text{m}/\text{s} \right) \\ \textbf{p}_{\text{tranquilizer}} & = 24\ \text{kg} \cdot \text{m}/\text{s} \end{align*}

Now, we can compare their momentums.

\begin{align*} \frac{\textbf{p}_{\text{elephant}}}{\textbf{p}_{\text{tranquilizer}}} & = \frac{15000\ \text{kg} \cdot \text{m}/\text{s}}{24\ \text{kg} \cdot \text{m}/\text{s} } \\ \frac{\textbf{p}_{\text{elephant}}}{\textbf{p}_{\text{tranquilizer}}} & = 625 \\ \textbf{p}_{\text{elephant}} & = 625\ \textbf{p}_{\text{tranquilizer}} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The momentum of the elephant is 625 times larger than the momentum of the tranquilizer.

Part C. The Momentum of a Hunter Running after missing the elephant

\begin{align*} \textbf{p}_{\text{hunter}} & = m_{\text{hunter}} \textbf{v}_{\text{hunter}} \\ \textbf{p}_{\text{hunter}} & = \left( 90.0\ \text{kg} \right)\left( 7.40\ \text{m}/\text{s} \right) \\ \textbf{p}_{\text{hunter}} & = 666\ \text{kg} \cdot \text{m}/\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Problem 6-8: An integrated problem involving circular motion, momentum, and projectile motion

Integrated Concepts

When kicking a football, the kicker rotates his leg about the hip joint.

(a) If the velocity of the tip of the kicker’s shoe is 35.0 m/s and the hip joint is 1.05 m from the tip of the shoe, what is the shoe tip’s angular velocity?

(b) The shoe is in contact with the initially stationary 0.500 kg football for 20.0 ms. What average force is exerted on the football to give it a velocity of 20.0 m/s?

(c) Find the maximum range of the football, neglecting air resistance.

Solution:

Part A

From the given problem, we are given the following values: $v=35.0\ \text{m/s}$ and $r=1.05\ \text{m}$ . We are required to solve for the angular velocity $\omega$ .

The linear velocity, $v$ and the angular velocity, $\omega$ are related by the equation

v=r\omega \ \text{or} \ \omega=\frac{v}{r}

If we substitute the given values into the formula, we can directly solve for the value of the angular velocity. That is,

\begin{align*} \omega & = \frac{v}{r} \\ \\ \omega & = \frac{35.0\ \text{m/s}}{1.05\ \text{m}} \\ \\ \omega & = 33.3333\ \text{rad/sec} \\ \\ \omega & = 33.3 \ \text{rad/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

For this part of the problem, we are going to use Newton’s second law of motion in term of linear momentum which states that the net external force equals the change in momentum of a system divided by the time over which it changes. That is

F_{net} = \frac{\Delta p}{\Delta t} = \frac{m\left( v_f - v_i \right)}{t}

For this problem, we are given the following values: $m=0.500\ \text{kg}$ , $t=20.0\times 10^{-3} \ \text{s}$ , $v_{f}=20.0\ \text{m/s}$ , and $v_{i}=0$ . Substituting all these values into the equation, we can solve directly for the value of the net external force.

\begin{align*} F_{net} & = \frac{\left( 0.500\ \text{kg} \right)\left( 20.0\ \text{m/s}-0\ \text{m/s} \right)}{20.0\times 10^{-3}\ \text{s}} \\ \\ F_{net} & = 500\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

This is a problem on projectile motion. In this particular case, we are solving for the range of the projectile. The formula for the range of a projectile is

R=\frac{v_{0}^2 \sin 2\theta}{g}

We are asked to solve for the maximum range, and we know that the maximum range happens when the angle $\theta$ is $45^\circ$ .

\begin{align*} R & = \frac{\left( 20.0\ \text{m/s} \right)^{2} \sin \left( 2\left( 45^\circ \right) \right)}{9.81 \ \text{m/s}^2} \\ \\ R & = 40.7747\ \text{m} \\ \\ R & = 40.8 \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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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

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

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A car moving at 10.0 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat belt exerts on a passenger in the car to bring him to a halt. The mass of the passenger is 70.0 kg.

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600.0 m/s in a time of 2.00 ms (milliseconds)?

The mass of Earth is 5.972×1024 kg5.972 \times 10^{24}\ \text{kg}5.972×1024 kg and its orbital radius is an average of 1.496×1011 m1.496 \times 10^{11}\ \text{m}1.496×1011 m. Calculate its linear momentum.

A runaway train car that has a mass of 15,000 kg travels at a speed of 5.4 m/s down a track. Compute the time required for a force of 1500 N to bring the car to rest.

(a) What is the momentum of a garbage truck that is 1.20×104 kg and is moving at 30.0 m/s? (b) At what speed would an 8.00-kg trash can have the same momentum as the truck?

(a) What is the mass of a large ship that has a momentum of 1.60×109 kg⋅m/s, when the ship is moving at a speed of 48.0 km/h? (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s.

Integrated Concepts

When kicking a football, the kicker rotates his leg about the hip joint.

(a) If the velocity of the tip of the kicker’s shoe is 35.0 m/s and the hip joint is 1.05 m from the tip of the shoe, what is the shoe tip’s angular velocity?

(b) The shoe is in contact with the initially stationary 0.500 kg football for 20.0 ms. What average force is exerted on the football to give it a velocity of 20.0 m/s?

(c) Find the maximum range of the football, neglecting air resistance.

Solution:

Part A

Part B

Part C

Linear Momentum and Force

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

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

Conservation of Momentum

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Elastic Collisions in One Dimension

Problem 28

Problem 29

Problem 30

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

Collisions of Point Masses in Two Dimensions

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Introduction to Rocket Propulsion

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

The mass of Earth is $5.972 \times 10^{24}\ \text{kg}$ and its orbital radius is an average of $1.496 \times 10^{11}\ \text{m}$ . Calculate its linear momentum.

(a) What is the momentum of a garbage truck that is 1.20×10⁴ kg and is moving at 30.0 m/s? (b) At what speed would an 8.00-kg trash can have the same momentum as the truck?

(a) What is the mass of a large ship that has a momentum of 1.60×10⁹ kg⋅m/s, when the ship is moving at a speed of 48.0 km/h? (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s.