A block of mass 2.0 kg moving towards right at a speed of 20.0 m/s collide elastically with another block of mass 4.0 kg moving towards left at a speed of 8.0 m/s. Calculate final speed of the blocks V₁f and V2f after collision.
A block of mass 2.0 kg moving towards right at a speed of 20.0 m/s collide elastically with another block of mass 4.0 kg moving towards left at a speed of 8.0 m/s. Calculate final speed of the blocks V₁f and V2f after collision.
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Author:Raymond A. Serway, Chris Vuille
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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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![**Problem Statement:**
A block of mass 2.0 kg moving towards the right at a speed of 20.0 m/s collides elastically with another block of mass 4.0 kg moving towards the left at a speed of 8.0 m/s. Calculate the final speed of the blocks \( V_{1f} \) and \( V_{2f} \) after the collision.
**Explanation:**
In this problem, we have two blocks moving towards each other with initial speeds. The collision is elastic, meaning both momentum and kinetic energy are conserved. You need to use these conservation principles to determine the final velocities of the blocks after collision.
You can start by applying the conservation of momentum equation:
\[ m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot V_{1f} + m_2 \cdot V_{2f} \]
And the conservation of kinetic energy equation:
\[ \frac{1}{2} m_1 \cdot v_{1i}^2 + \frac{1}{2} m_2 \cdot v_{2i}^2 = \frac{1}{2} m_1 \cdot V_{1f}^2 + \frac{1}{2} m_2 \cdot V_{2f}^2 \]
Where:
- \( m_1 \) = 2.0 kg, \( v_{1i} = 20.0 \) m/s (initial velocity of block 1)
- \( m_2 \) = 4.0 kg, \( v_{2i} = -8.0 \) m/s (initial velocity of block 2, negative because it's moving in the opposite direction)
- \( V_{1f} \) and \( V_{2f} \) are the final velocities to be determined.
Solve this system of equations to find the values of \( V_{1f} \) and \( V_{2f} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F55141c12-66e7-4047-bdc9-a7e626131e3b%2F71bde6d6-8264-486e-97b4-3f39cf97c04a%2Fenwy8ef_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A block of mass 2.0 kg moving towards the right at a speed of 20.0 m/s collides elastically with another block of mass 4.0 kg moving towards the left at a speed of 8.0 m/s. Calculate the final speed of the blocks \( V_{1f} \) and \( V_{2f} \) after the collision.
**Explanation:**
In this problem, we have two blocks moving towards each other with initial speeds. The collision is elastic, meaning both momentum and kinetic energy are conserved. You need to use these conservation principles to determine the final velocities of the blocks after collision.
You can start by applying the conservation of momentum equation:
\[ m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot V_{1f} + m_2 \cdot V_{2f} \]
And the conservation of kinetic energy equation:
\[ \frac{1}{2} m_1 \cdot v_{1i}^2 + \frac{1}{2} m_2 \cdot v_{2i}^2 = \frac{1}{2} m_1 \cdot V_{1f}^2 + \frac{1}{2} m_2 \cdot V_{2f}^2 \]
Where:
- \( m_1 \) = 2.0 kg, \( v_{1i} = 20.0 \) m/s (initial velocity of block 1)
- \( m_2 \) = 4.0 kg, \( v_{2i} = -8.0 \) m/s (initial velocity of block 2, negative because it's moving in the opposite direction)
- \( V_{1f} \) and \( V_{2f} \) are the final velocities to be determined.
Solve this system of equations to find the values of \( V_{1f} \) and \( V_{2f} \).
Expert Solution
Step 1
Given:
The mass of the first block is 2 kg.
The initial speed of the first block is 20 m/s in the right direction.
The mass of the second block is 4 kg.
The initial speed of the second block is 8 m/s in the left direction.
The collision is perfectly elastic.
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