Tag Archives: Impulse

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 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.

$Solution Guides to College Physics by Openstax Chapter 8 Banner$

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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