Two cars traveling in opposite directions collide and stick together. Before colliding, the first (A) has a mass of 21 kg and travels with a velocity of 7 m/s while the second (B) has a mass of 15 kg and a velocity of -1 m/s. After colliding, what is the velocity of the pair? - V=?
Two cars traveling in opposite directions collide and stick together. Before colliding, the first (A) has a mass of 21 kg and travels with a velocity of 7 m/s while the second (B) has a mass of 15 kg and a velocity of -1 m/s. After colliding, what is the velocity of the pair? - V=?
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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![**Educational Content: Momentum and Collisions**
**Scenario:**
Two cars traveling in opposite directions collide and stick together. Before colliding:
- **Car A:**
- Mass = 21 kg
- Velocity = 7 m/s (to the right)
- **Car B:**
- Mass = 15 kg
- Velocity = -1 m/s (to the left)
**Question:**
After colliding, what is the velocity of the combined pair?
**Diagram Explanation:**
1. **Initial State (Top Diagram):**
- Two separate cars (A and B) are shown on tracks.
- Car A moves to the right with a velocity of 7 m/s.
- Car B moves to the left with a velocity of -1 m/s.
2. **Post-Collision State (Bottom Diagram):**
- The combined object (cars A and B stuck together) is on a track.
- The velocity of the combined object is denoted as V (unknown).
**Solution Approach:**
To solve this, use the principle of conservation of momentum, which states that the total momentum before the collision equals the total momentum after the collision.
**Formula:**
\[ (m_A \cdot v_A) + (m_B \cdot v_B) = (m_A + m_B) \cdot V \]
**Calculations:**
\[ (21 \, \text{kg} \cdot 7 \, \text{m/s}) + (15 \, \text{kg} \cdot -1 \, \text{m/s}) = (21 \, \text{kg} + 15 \, \text{kg}) \cdot V \]
\[ 147 \, \text{kg·m/s} - 15 \, \text{kg·m/s} = 36 \, \text{kg} \cdot V \]
\[ 132 \, \text{kg·m/s} = 36 \, \text{kg} \cdot V \]
\[ V = \frac{132 \, \text{kg·m/s}}{36 \, \text{kg}} \]
\[ V = 3.67 \, \text{m/s} \]
**Conclusion:**
The velocity of the combined pair after the collision is approximately 3.67 m/s to the right.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a857556-438b-4e88-a66d-cf80345c3048%2Fa49f4405-e7d3-43bf-8d0a-737ff5a8601e%2Fd1i6jf_processed.png&w=3840&q=75)
Transcribed Image Text:**Educational Content: Momentum and Collisions**
**Scenario:**
Two cars traveling in opposite directions collide and stick together. Before colliding:
- **Car A:**
- Mass = 21 kg
- Velocity = 7 m/s (to the right)
- **Car B:**
- Mass = 15 kg
- Velocity = -1 m/s (to the left)
**Question:**
After colliding, what is the velocity of the combined pair?
**Diagram Explanation:**
1. **Initial State (Top Diagram):**
- Two separate cars (A and B) are shown on tracks.
- Car A moves to the right with a velocity of 7 m/s.
- Car B moves to the left with a velocity of -1 m/s.
2. **Post-Collision State (Bottom Diagram):**
- The combined object (cars A and B stuck together) is on a track.
- The velocity of the combined object is denoted as V (unknown).
**Solution Approach:**
To solve this, use the principle of conservation of momentum, which states that the total momentum before the collision equals the total momentum after the collision.
**Formula:**
\[ (m_A \cdot v_A) + (m_B \cdot v_B) = (m_A + m_B) \cdot V \]
**Calculations:**
\[ (21 \, \text{kg} \cdot 7 \, \text{m/s}) + (15 \, \text{kg} \cdot -1 \, \text{m/s}) = (21 \, \text{kg} + 15 \, \text{kg}) \cdot V \]
\[ 147 \, \text{kg·m/s} - 15 \, \text{kg·m/s} = 36 \, \text{kg} \cdot V \]
\[ 132 \, \text{kg·m/s} = 36 \, \text{kg} \cdot V \]
\[ V = \frac{132 \, \text{kg·m/s}}{36 \, \text{kg}} \]
\[ V = 3.67 \, \text{m/s} \]
**Conclusion:**
The velocity of the combined pair after the collision is approximately 3.67 m/s to the right.
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