bottom! 1. A ball is dropped from a height of 1 m and loses 10 percent of its kinetic energy when it bounces on the ground. To what height does it then rise? alm is struck by a

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)...
icon
Related questions
icon
Concept explainers
Topic Video
Question
**Physics Problem: Ball Drop and Energy Loss**

**Problem Statement:**
A ball is dropped from a height of 1 meter and loses 10 percent of its kinetic energy when it bounces on the ground. To what height does it then rise?

---

**Solution Approach:**

1. **Initial Potential Energy (PE):**
   - The ball is dropped from a height of \( h = 1 \) meter.
   - Potential Energy at height \( h \) is given by \( PE = mgh \), where \( m \) is the mass of the ball, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and \( h \) is the height.

2. **Conversion to Kinetic Energy:**
   - When the ball is at its maximum height, the potential energy is fully converted to kinetic energy just before it hits the ground (ignoring air resistance).
   - Therefore, the kinetic energy \( KE \) just before impact is equal to the initial potential energy, i.e., \( KE = mgh \).

3. **Energy Loss on Impact:**
   - The ball loses 10 percent of its kinetic energy upon impact.
   - Remaining kinetic energy after the bounce = \( 0.9 \times KE \).

4. **Conversion Back to Potential Energy:**
   - After the bounce, the remaining kinetic energy is converted back into potential energy at the new height \( h_{\text{new}} \).
   - Therefore, \( 0.9 \times mgh = mgh_{\text{new}} \).

5. **Solving for New Height \( h_{\text{new}} \):**
   - Cancel the mass \( m \) from both sides of the equation:
     \[
     0.9 \times gh = gh_{\text{new}}
     \]
   - Simplify to find:
     \[
     h_{\text{new}} = 0.9 \times h
     \]
   - Since \( h = 1 \) meter initially:
     \[
     h_{\text{new}} = 0.9 \times 1 \text{ meter} = 0.9 \text{ meters}
     \]

**Therefore, the ball rises to a height of 0.9 meters after bouncing.**
Transcribed Image Text:**Physics Problem: Ball Drop and Energy Loss** **Problem Statement:** A ball is dropped from a height of 1 meter and loses 10 percent of its kinetic energy when it bounces on the ground. To what height does it then rise? --- **Solution Approach:** 1. **Initial Potential Energy (PE):** - The ball is dropped from a height of \( h = 1 \) meter. - Potential Energy at height \( h \) is given by \( PE = mgh \), where \( m \) is the mass of the ball, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and \( h \) is the height. 2. **Conversion to Kinetic Energy:** - When the ball is at its maximum height, the potential energy is fully converted to kinetic energy just before it hits the ground (ignoring air resistance). - Therefore, the kinetic energy \( KE \) just before impact is equal to the initial potential energy, i.e., \( KE = mgh \). 3. **Energy Loss on Impact:** - The ball loses 10 percent of its kinetic energy upon impact. - Remaining kinetic energy after the bounce = \( 0.9 \times KE \). 4. **Conversion Back to Potential Energy:** - After the bounce, the remaining kinetic energy is converted back into potential energy at the new height \( h_{\text{new}} \). - Therefore, \( 0.9 \times mgh = mgh_{\text{new}} \). 5. **Solving for New Height \( h_{\text{new}} \):** - Cancel the mass \( m \) from both sides of the equation: \[ 0.9 \times gh = gh_{\text{new}} \] - Simplify to find: \[ h_{\text{new}} = 0.9 \times h \] - Since \( h = 1 \) meter initially: \[ h_{\text{new}} = 0.9 \times 1 \text{ meter} = 0.9 \text{ meters} \] **Therefore, the ball rises to a height of 0.9 meters after bouncing.**
Expert Solution
steps

Step by step

Solved in 3 steps with 3 images

Blurred answer
Knowledge Booster
Kinetic energy
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON