Tag Archives: College Physics by Openstax Solution Manual

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*}

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*}

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*}

College Physics by Openstax Chapter 7 Problem 10

(a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates.

Solution:

The translational kinetic energy of an object of mass $m$ moving at speed $v$ is $KE=\frac{1}{2}mv^{2}$ .

Part A. The Velocity of the Elephant to have the same Kinetic Energy as the Sprinter

First, we need to solve for the kinetic energy of the sprinter.

\begin{align*} KE_{\text{sprinter}} & = \frac{1}{2} \left( 65.0\ \text{kg} \right)\left( 10.0\ \text{m}/\text{s} \right)^{2} \\ KE_{\text{sprinter}} & = 3250\ \text{J} \end{align*}

Then, we need to equate this to the kinetic energy of the elephant with the velocity as the unknown.

\begin{align*} KE_{\text{elephant}} & = KE_{\text{sprinter}} \\ \frac{1}{2}\left( 3000\ \text{kg} \right) v^{2} & = 3250\ \text{J} \\ 1500 v^{2} & = 3250 \\ v^{2} & = \frac{3250}{1500} \\ v & = \sqrt[]{\frac{3250}{1500}} \\ v & = 1.4720\ \text{m}/\text{s} \\ v & = 1.47\ \text{m}/\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. How Larger Energies Needed for the Movement of Larger Animals would Relate to Metabolic Rates

If the elephant and the sprinter accelerate to a final velocity of 10.0 m/s, then
the elephant would have a much larger kinetic energy than the sprinter.
Therefore, the elephant clearly has burned more energy and requires a faster
metabolic output to sustain that speed. $\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)$

College Physics by Openstax Chapter 7 Problem 9

Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 km/h.

Solution:

The translational kinetic energy of an object of mass $m$ moving at speed $v$ is $KE=\frac{1}{2}mv^{2}$ .

The Kinetic Energy of the Truck

For the truck, we are given the following:

\begin{align*} m & = 20 000\ \text{kg} \\ v & = 110\ \frac{\text{km}}{\text{hr}} \times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{hr}}{3600\ \text{s}} = 30.5556\ \text{m}/\text{s} \end{align*}

Substitute these values to compute for the kinetic energy of the truck.

\begin{align*} KE_{t} & = \frac{1}{2} mv^{2} \\ KE_{t} & = \frac{1}{2} \left( 20 000\ \text{kg} \right) \left( 30.5556\ \text{m}/\text{s} \right)^{2} \\ KE_{t} & = 9 336 446.9136\ \text{J} \\ KE_{t} & = 9.34 \times 10^{6} \ \text{J} \end{align*}

The Kinetic Energy of the Astronaut

For the astronaut, we have the following given values

\begin{align*} m & = 80\ \text{kg} \\ v & = 27 500\ \frac{\text{km}}{\text{hr}} \times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{hr}}{3600\ \text{s}} = 7638.8889\ \text{m}/\text{s} \end{align*}

The kinetic energy of the astronaut is calculated as

\begin{align*} KE_{a} & = \frac{1}{2} mv^{2} \\ KE_{a} & = \frac{1}{2} \left( 80\ \text{kg} \right) \left( 7638.8889\ \text{m}/\text{s} \right)^{2} \\ KE_{a} & = 2 334 104 945 .0617\ \text{J} \\ KE_{a} & = 2.33 \times 10^{9} \ \text{J} \end{align*}

Comparing the Kinetic Energies of the truck and the astronaut

\begin{align*} \frac{KE_{a}}{KE_{t}} & = \frac{2 334 104 945 .0617\ \text{J}}{9 336 446.9136\ \text{J}} \\ \frac{KE_{a}}{KE_{t}} & = 250 \\ KE_{a} & = 250\ KE_{t} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The kinetic energy of the astronaut is 250 times larger than the kinetic energy of the truck.

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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.

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)?

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.

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.

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?

College Physics by Openstax Chapter 8 Problem 3

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.

College Physics by Openstax Chapter 8 Problem 1

College Physics by Openstax Chapter 7 Problem 10

(a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates.

College Physics by Openstax Chapter 7 Problem 9

Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 km/h.

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.

(a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates.

Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 km/h.

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.