### Problem DescriptionIn the figure below, block 1 of mass \( m_1 \) slides from rest along a frictionless ramp from height \( h = 2.50 \, \text{m} \) and then collides with stationary block 2, which has mass \( m_2 = 2.00m_1 \). After the collision, block 2 slides into a region where the coefficient of kinetic friction \( \mu_k \) is 0.500 and comes to a stop in distance \( d \) within that region. What is the value of distance \( d \) if the collision is:(a) Elastic(b) Completely inelastic?### Diagram DescriptionThe diagram associated with the problem includes the following elements:- **Frictionless Ramp**: The ramp is curved and smooth, allowing block 1 to slide without any friction.- **Block 1**: Positioned at the top of the frictionless ramp with height \( h \).- **Block 2**: Initially stationary and situated at the bottom of the frictionless ramp.- **Kinetic Friction Region**: After collision, block 2 encounters a rough surface with a kinetic friction coefficient \( \mu_k = 0.500 \).### Analyzing the ProblemTo solve this problem, we need to address two scenarios: elastic collision and completely inelastic collision.#### (a) **Elastic Collision**1. **Initial Potential Energy (PE) of Block 1**: \[ PE = m_1gh \]2. **Kinetic Energy (KE) at Bottom of Ramp**: \[ KE = \frac{1}{2}m_1v_1^2 \]\[ m_1gh = \frac{1}{2}m_1v_1^2 \rightarrow v_1 = \sqrt{2gh} \]3. **Velocity of Block 1 before Collision**:\[ v_1 = \sqrt{2 \cdot 9.8 \cdot 2.50} \approx 7 \, \text{m/s} \]4. **Velocity of Both Blocks after Elastic Collision**:For an elastic collision:\[ v_{1f} = \frac{(m_1 - m_2)}{(m_1 + m_2)}v_1 \quad \text{and} \quad v_{2

Answered: Image /qna-images/answer/31eb9e20-7433-4f1e-885b-a7ab8875a949.jpg

= (b) In the figure below, block 1 of mass m₁ slides from rest along a frictionless ramp from height h 2.50 m and then collides with stationary block 2, which has mass m2 = 2.00m₁. After the collision, block 2 slides into a region where the coefficient of kinetic friction k is 0.500 and comes to a stop in distance d within that region. What is the value of distance d if the collision is (a) elastic and (b) completely inelastic? Frictionless 2

= (b) In the figure below, block 1 of mass m₁ slides from rest along a frictionless ramp from height h 2.50 m and then collides with stationary block 2, which has mass m2 = 2.00m₁. After the collision, block 2 slides into a region where the coefficient of kinetic friction k is 0.500 and comes to a stop in distance d within that region. What is the value of distance d if the collision is (a) elastic and (b) completely inelastic? Frictionless 2

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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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Question

### Problem Description

In the figure below, block 1 of mass \( m_1 \) slides from rest along a frictionless ramp from height \( h = 2.50 \, \text{m} \) and then collides with stationary block 2, which has mass \( m_2 = 2.00m_1 \). After the collision, block 2 slides into a region where the coefficient of kinetic friction \( \mu_k \) is 0.500 and comes to a stop in distance \( d \) within that region. What is the value of distance \( d \) if the collision is:
(a) Elastic
(b) Completely inelastic?

### Diagram Description

The diagram associated with the problem includes the following elements:

- **Frictionless Ramp**: The ramp is curved and smooth, allowing block 1 to slide without any friction.
- **Block 1**: Positioned at the top of the frictionless ramp with height \( h \).
- **Block 2**: Initially stationary and situated at the bottom of the frictionless ramp.
- **Kinetic Friction Region**: After collision, block 2 encounters a rough surface with a kinetic friction coefficient \( \mu_k = 0.500 \).

### Analyzing the Problem

To solve this problem, we need to address two scenarios: elastic collision and completely inelastic collision.
#### (a) **Elastic Collision**

1. **Initial Potential Energy (PE) of Block 1**:
\[ PE = m_1gh \]
2. **Kinetic Energy (KE) at Bottom of Ramp**:
\[ KE = \frac{1}{2}m_1v_1^2 \]
\[ m_1gh = \frac{1}{2}m_1v_1^2 \rightarrow v_1 = \sqrt{2gh} \]
3. **Velocity of Block 1 before Collision**:
\[ v_1 = \sqrt{2 \cdot 9.8 \cdot 2.50} \approx 7 \, \text{m/s} \]
4. **Velocity of Both Blocks after Elastic Collision**:
For an elastic collision:
\[ v_{1f} = \frac{(m_1 - m_2)}{(m_1 + m_2)}v_1 \quad \text{and} \quad v_{2

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