College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Understanding Impulse and Momentum
Impulse and momentum are fundamental concepts in physics that describe the motion of objects. Here's a breakdown of the key principles and equations involved:
#### Key Equations:
1. **Momentum (p)**
\[
p = mv
\]
* \( p \) = momentum
* \( m \) = mass (in kilograms)
* \( v \) = velocity (in meters per second)
2. **Impulse (J)**
\[
J = Ft
\]
* \( J \) = impulse
* \( F \) = force applied (in Newtons)
* \( t \) = time over which the force is applied (in seconds)
3. **Impulse-Momentum Theorem**
\[
Ft = m \Delta v
\]
* \( F \) = force
* \( t \) = time
* \( m \) = mass
* \( \Delta v \) = change in velocity
#### Problem Statement
What impulse is needed to stop a 1700 kg car in 15 seconds if it was originally moving at 25 m/s?
To solve this problem, we need to find the impulse (\( J \)) required to bring the car to a stop. We are given:
- Mass of the car (\( m \)) = 1700 kg
- Initial velocity (\( v \)) = 25 m/s
- Final velocity (\( v_f \)) = 0 m/s (since the car stops)
- Time (\( t \)) = 15 seconds
Using the Impulse-Momentum Theorem (\( Ft = m \Delta v \)), we can calculate the impulse.
First, determine the change in velocity (\( \Delta v \)):
\[
\Delta v = v_f - v = 0 - 25 = -25 \, \text{m/s}
\]
Then, calculate the impulse:
\[
J = m \Delta v
\]
\[
J = 1700 \times (-25)
\]
\[
J = -42500 \, \text{Ns}
\]
The negative sign indicates that the impulse is in the direction opposite to the car's initial motion in order to bring it to a stop. Thus, the magnitude of the impulse needed is **42500 Ns**.
This example illustrates how to apply the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbb95dca4-f6c7-46fa-bbcb-5cdbe2ca0e09%2F09a18b58-52aa-4bf1-9fa3-d223506e416f%2Fcoha3pd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding Impulse and Momentum
Impulse and momentum are fundamental concepts in physics that describe the motion of objects. Here's a breakdown of the key principles and equations involved:
#### Key Equations:
1. **Momentum (p)**
\[
p = mv
\]
* \( p \) = momentum
* \( m \) = mass (in kilograms)
* \( v \) = velocity (in meters per second)
2. **Impulse (J)**
\[
J = Ft
\]
* \( J \) = impulse
* \( F \) = force applied (in Newtons)
* \( t \) = time over which the force is applied (in seconds)
3. **Impulse-Momentum Theorem**
\[
Ft = m \Delta v
\]
* \( F \) = force
* \( t \) = time
* \( m \) = mass
* \( \Delta v \) = change in velocity
#### Problem Statement
What impulse is needed to stop a 1700 kg car in 15 seconds if it was originally moving at 25 m/s?
To solve this problem, we need to find the impulse (\( J \)) required to bring the car to a stop. We are given:
- Mass of the car (\( m \)) = 1700 kg
- Initial velocity (\( v \)) = 25 m/s
- Final velocity (\( v_f \)) = 0 m/s (since the car stops)
- Time (\( t \)) = 15 seconds
Using the Impulse-Momentum Theorem (\( Ft = m \Delta v \)), we can calculate the impulse.
First, determine the change in velocity (\( \Delta v \)):
\[
\Delta v = v_f - v = 0 - 25 = -25 \, \text{m/s}
\]
Then, calculate the impulse:
\[
J = m \Delta v
\]
\[
J = 1700 \times (-25)
\]
\[
J = -42500 \, \text{Ns}
\]
The negative sign indicates that the impulse is in the direction opposite to the car's initial motion in order to bring it to a stop. Thus, the magnitude of the impulse needed is **42500 Ns**.
This example illustrates how to apply the
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