Lesson Plan | Traditional Methodology | Momentum and Impulse: Coefficient of Restitution

Objectives

Duration: 10 - 15 minutes

The purpose of this stage is to establish a solid foundation on what is expected for the students to learn during the lesson. Defining the main objectives helps to direct the focus of the explanation and ensures that students know exactly what they should comprehend and be capable of doing by the end of the lesson.

Main Objectives

1. Understand what the coefficient of restitution is and its meaning.

2. Identify and differentiate the different types of collisions.

3. Apply the coefficient of restitution to calculate velocity values before or after collisions.

Introduction

Duration: 10 - 15 minutes

 The purpose of this stage is to create an engaging and relevant starting point for the lesson. By providing an initial context and curiosities, students are introduced to the topic in an interesting and practical way, which increases engagement and understanding. This introduction establishes the foundation for the detailed explanations that will follow, helping to solidify the importance of the coefficient of restitution in collisions.

Context

 Start the lesson by contextualizing the students on the topic of the day: Impulse and Momentum with a focus on the Coefficient of Restitution. Explain that in Physics, the study of collisions and impacts is essential to understand how bodies interact with each other. Momentum (linear momentum) and impulse are concepts that help describe these interactions. The coefficient of restitution, in turn, is a measure of the 'elasticity' of a collision, meaning how well kinetic energy is conserved in an impact. It is crucial for predicting the post-collision behavior of the bodies involved.

Curiosities

 To engage the students, share a curiosity: Did you know that the coefficient of restitution is a concept used in various sports? For example, in the manufacturing of basketball and tennis balls, it is essential that they have the correct coefficient of restitution to ensure they bounce adequately. Additionally, it is used in studies of car accidents to understand the dynamics of collisions and improve vehicle safety.

Development

Duration: 50 - 60 minutes

 The purpose of this stage is to deepen the students' understanding of the coefficient of restitution through detailed explanations and practical examples. By addressing the different types of collisions and solving applied problems, students will have the opportunity to apply theoretical concepts in practical situations, thus consolidating their learning and preparing to use this information in various contexts.

Covered Topics

1.  Definition of Coefficient of Restitution (COR): Explain that the coefficient of restitution is a measure of the 'elasticity' of a collision, defined as the ratio of the relative velocity of separation to the relative velocity of approach of the bodies after and before the collision, respectively. The formula is: COR = (v2' - v1') / (v1 - v2), where v1 and v2 are the velocities before the collision and v1' and v2' are the velocities after the collision. 2. ⚙️ Types of Collisions: Detail the different types of collisions based on the coefficient of restitution. Explain that a collision is perfectly elastic when COR = 1, partially elastic when 0 < COR < 1, and perfectly inelastic when COR = 0. 3.  Practical Examples: Provide practical examples and solve guided problems on how to calculate the velocities of bodies before and after collisions using the coefficient of restitution. For example, if two billiard balls collide, how can you determine their velocities after the collision?

Classroom Questions

1. Calculate the coefficient of restitution if, after a collision, an object moving at 5 m/s rebounds with a speed of 3 m/s and the second object, initially at rest, moves with a speed of 2 m/s. 2. Two balls collide head-on. The first ball, with a mass of 2 kg, moves at 4 m/s, while the second ball, with a mass of 3 kg, moves at -2 m/s. After the collision, the first ball moves at 1 m/s. Calculate the speed of the second ball and the coefficient of restitution. 3. In a perfectly inelastic collision, two cars collide and move together after the collision. If the first car had a speed of 10 m/s and the second car, initially at rest, has a mass equal to the first car, what will be the speed of the cars after the collision?

Questions Discussion

Duration: 25 - 30 minutes

 The purpose of this stage is to review and consolidate students' learning through the discussion of the resolved questions and reflection on the presented concepts. By discussing the answers and raising additional questions, students have the opportunity to deepen their understanding, correct possible mistakes, and connect theoretical concepts with practical and everyday applications.

Discussion

•  Question 1: Calculate the coefficient of restitution if, after a collision, an object moving at 5 m/s rebounds with a speed of 3 m/s and the second object, initially at rest, moves with a speed of 2 m/s.

• Explanation: To calculate the coefficient of restitution (COR), use the formula COR = (v2' - v1') / (v1 - v2). In this case, v1 = 5 m/s, v2 = 0 m/s, v1' = 3 m/s, and v2' = 2 m/s. Therefore, COR = (2 - 3) / (5 - 0) = -1 / 5 = -0.2. As the coefficient of restitution is an absolute value, COR is 0.2.

•  Question 2: Two balls collide head-on. The first ball, with a mass of 2 kg, moves at 4 m/s, while the second ball, with a mass of 3 kg, moves at -2 m/s. After the collision, the first ball moves at 1 m/s. Calculate the speed of the second ball and the coefficient of restitution.

• Explanation: Use the conservation of momentum to find the speed of the second ball. Before the collision: (2 kg * 4 m/s) + (3 kg * -2 m/s) = 8 kgm/s - 6 kgm/s = 2 kgm/s. After the collision: (2 kg * 1 m/s) + (3 kg * v2') = 2 kgm/s + 3 kg * v2'. Equating the two expressions: 2 kgm/s = 2 kgm/s + 3 kg * v2', solving for v2', we find v2' = 0 m/s.

• For the COR, use the formula: COR = (v2' - v1') / (v1 - v2). In this case, v1 = 4 m/s, v2 = -2 m/s, v1' = 1 m/s, and v2' = 0 m/s. Therefore, COR = (0 - 1) / (4 - (-2)) = -1 / 6 ≈ -0.17. COR is 0.17.

•  Question 3: In a perfectly inelastic collision, two cars collide and move together after the collision. If the first car had a speed of 10 m/s and the second car, initially at rest, has a mass equal to the first car, what will be the speed of the cars after the collision?

• Explanation: In a perfectly inelastic collision, the bodies move together after the collision. Using the conservation of momentum: (m * 10 m/s) + (m * 0 m/s) = (2m) * v'. Simplifying, we have: 10m = 2m * v', solving for v', we find v' = 5 m/s.

Student Engagement

1. 樂 Question: How does the coefficient of restitution influence the behavior of balls in different sports? 2. 樂 Question: Why is it important to consider the coefficient of restitution in automotive safety studies? 3. 樂 Question: If a collision has a coefficient of restitution close to 1, what does that indicate about the nature of the collision? 4. 樂 Reflection: How does the conservation of momentum relate to the coefficient of restitution in different types of collisions? 5. 樂 Reflection: In what other everyday situations can we observe the application of the coefficient of restitution?

Conclusion

Duration: 10 - 15 minutes

The purpose of this stage is to review and consolidate the knowledge acquired by the students, providing a summary of the main points discussed, highlighting the connection between theory and practice, and emphasizing the relevance of the topic to everyday life. This recap helps ensure that students have a clear and lasting understanding of the concepts presented.

Summary

• Definition of the coefficient of restitution (COR) as the ratio between the relative velocity of separation and the relative velocity of approach of the bodies after and before the collision.
• Types of collisions: perfectly elastic (COR = 1), partially elastic (0 < COR < 1), and perfectly inelastic (COR = 0).
• Practical examples of calculating the coefficient of restitution and the velocities of bodies before and after collisions.
• Application of momentum conservation in collision problems.

The lesson connected theory with practice by thoroughly explaining the concept of the coefficient of restitution and demonstrating its application in various practical examples, such as analyzing collisions between billiard balls and cars, as well as highlighting its importance in contexts such as sports and automotive safety.

The coefficient of restitution is a crucial concept for understanding the dynamics of collisions, with practical applications in areas such as vehicle safety engineering and the manufacturing of sports equipment. Its understanding allows for the prediction and analysis of the behavior of objects after impacts, becoming essential for improving safety and efficiency in various everyday situations.