### Problem StatementA 2 kg block is moving at 5 m/s along a frictionless table and collides with a second 2 kg block that is initially at rest. After the collision, the two blocks stick together and then slide up a 45° frictionless inclined plane, as shown in the diagram below. Calculate the maximum distance \( L \) that the two blocks travel up the incline.### Diagram ExplanationThe diagram illustrates two stages of the blocks' movement:1. **Pre-Collision Movement:** - A 2 kg block is shown moving horizontally to the right at a speed of 5 m/s. - A second 2 kg block is shown at rest on the same frictionless surface.2. **Post-Collision Movement:** - After the collision, the two blocks stick together and move as one unit. - This combined block then slides up a frictionless inclined plane that makes a 45° angle with the horizontal surface. - The length \( L \) represents the maximum distance the blocks travel up the incline.### Key Concepts for Solution- Conservation of Momentum: Since the surface is frictionless, momentum is conserved during the collision.- Energy Conservation: The kinetic energy of the blocks is converted into gravitational potential energy as they move up the incline.### Steps to Solution1. **Calculate the velocity after collision using conservation of momentum:** \[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \] where \( m_1 = 2 \, \text{kg} \), \( v_1 = 5 \, \text{m/s} \), \( m_2 = 2 \, \text{kg} \), and \( v_2 = 0 \, \text{m/s} \).2. **Convert the kinetic energy of the combined block after collision into gravitational potential energy to find the maximum height \( h \):** \[ \frac{1}{2} (m_1 + m_2) v_f^2 = (m_1 + m_2) g h \] where \( g = 9.8 \, \text{m/s}^2 \).3. **Relate the height \( h \) to the distance \( L \) up the incline using trigon

Name: ### Problem StatementA 2 kg block is moving at 5 m/s along a friction
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Description: ### Problem StatementA 2 kg block is moving at 5 m/s along a frictionless table and collides with a second 2 kg block that is initially at rest. After the collision, the two blocks stick together and then slide up a 45° frictionless inclined plane,

Write the given values with the suitable variables. m=2 kgu1=5 m/su2=0 m/sθ=45° Here, m, u1, u2, and…

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(b) A 2 kg block is moving at 5 m/s along a frictionless table and collides with a second 2 kg block that is initially at rest. After the collision, the two blocks stick together and then slide up a 45° frictionless inclined plane, as shown in below. Calculate the maximum distance L that the two blocks travel up the incline. 5 m/s = 45°

(b) A 2 kg block is moving at 5 m/s along a frictionless table and collides with a second 2 kg block that is initially at rest. After the collision, the two blocks stick together and then slide up a 45° frictionless inclined plane, as shown in below. Calculate the maximum distance L that the two blocks travel up the incline. 5 m/s = 45°

College Physics

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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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### Problem Statement

A 2 kg block is moving at 5 m/s along a frictionless table and collides with a second 2 kg block that is initially at rest. After the collision, the two blocks stick together and then slide up a 45° frictionless inclined plane, as shown in the diagram below. Calculate the maximum distance \( L \) that the two blocks travel up the incline.

### Diagram Explanation

The diagram illustrates two stages of the blocks' movement:

1. **Pre-Collision Movement:**
- A 2 kg block is shown moving horizontally to the right at a speed of 5 m/s.
- A second 2 kg block is shown at rest on the same frictionless surface.

2. **Post-Collision Movement:**
- After the collision, the two blocks stick together and move as one unit.
- This combined block then slides up a frictionless inclined plane that makes a 45° angle with the horizontal surface.
- The length \( L \) represents the maximum distance the blocks travel up the incline.

### Key Concepts for Solution
- Conservation of Momentum: Since the surface is frictionless, momentum is conserved during the collision.
- Energy Conservation: The kinetic energy of the blocks is converted into gravitational potential energy as they move up the incline.

### Steps to Solution

1. **Calculate the velocity after collision using conservation of momentum:**
\[
m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f
\]
where \( m_1 = 2 \, \text{kg} \), \( v_1 = 5 \, \text{m/s} \), \( m_2 = 2 \, \text{kg} \), and \( v_2 = 0 \, \text{m/s} \).

2. **Convert the kinetic energy of the combined block after collision into gravitational potential energy to find the maximum height \( h \):**
\[
\frac{1}{2} (m_1 + m_2) v_f^2 = (m_1 + m_2) g h
\]
where \( g = 9.8 \, \text{m/s}^2 \).

3. **Relate the height \( h \) to the distance \( L \) up the incline using trigon

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