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Let's break down and solve this problem step by step:

### Problem Understanding:

- **Block 1** (mass = 1.60 kg) is moving at a velocity of \( v_1 = 3.60 \, \text{m/s} \).
- **Block 2** (mass = 1.30 kg) is at rest and touching an unstretched spring (spring constant \( k = 296 \, \text{N/m} \)) on a frictionless surface.
- After Block 1 collides with Block 2, they stick together and compress the spring. 
- We need to find the distance by which the spring is compressed when the blocks momentarily stop.

### Step-by-Step Solution:

1. **Conservation of Momentum (before and after collision):**

   The total momentum before the collision is the momentum of Block 1 (since Block 2 is at rest):

   \[
   p_{\text{initial}} = m_1 \cdot v_1 + m_2 \cdot v_2 = 1.60 \cdot 3.60 + 1.30 \cdot 0 = 5.76 \, \text{kg} \cdot \text{m/s}
   \]

   After the collision, the blocks stick together and move as a single mass \( (m_1 + m_2) \). Let \( v_f \) be their final velocity just after the collision but before the spring is compressed. Using conservation of momentum:

   \[
   m_1 \cdot v_1 + m_2 \cdot v_2 = (m_1 + m_2) \cdot v_f
   \]

   Solving for \( v_f \):

   \[
   v_f = \frac{5.76}{1.60 + 1.30} = \frac{5.76}{2.90} = 1.986 \, \text{m/s}
   \]

2. **Conservation of Energy:**

   Once the blocks stick together and compress the spring, the kinetic energy of the blocks is converted into potential energy stored in the spring.

   The total kinetic energy of the two blocks just after the collision is:

   \[
   K = \frac{1}{2} (m_1 + m_2) v_f^2
   \]

   Substituting \( v_f = 1.986 \, \text{m/s} \) and \( m_1 + m_2 = 2.90 \, \text{kg} \):

   \[
   K = \frac{1}{2} \cdot 2.90 \cdot (1.986)^2 = 2.87 \, \text{J}
   \]

   The potential energy stored in the spring when compressed by a distance \( x \) is:

   \[
   U = \frac{1}{2} k x^2
   \]

   Since all the kinetic energy is converted into spring potential energy:

   \[
   \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} k x^2
   \]

   Solving for \( x \):

   \[
   2.87 = \frac{1}{2} \cdot 296 \cdot x^2
   \]

   \[
   x^2 = \frac{2 \cdot 2.87}{296} = 0.01939
   \]

   \[
   x = \sqrt{0.01939} = 0.1393 \, \text{m}
   \]

### Final Answer:
The spring is compressed by \( \boxed{0.139} \, \text{m} \).

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Would you like further details or explanations on any part of the solution?

Here are 5 related questions for further practice:
1. What would happen if the surface was not frictionless?
2. How would the compression of the spring change if the spring constant was doubled?
3. What is the final velocity of the blocks before they stop?
4. How much energy is lost during the inelastic collision between the blocks?
5. How does the problem change if the blocks do not stick together?

**Tip:** When dealing with inelastic collisions, momentum is always conserved, but kinetic energy is not. Always check if energy transforms into another form (e.g., spring potential energy).

In the figure, block 2 (mass 1.30 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 296 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 1.60 kg), traveling at speed v1 = 3.60 m/s, collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?

Learn how to find the distance a spring is compressed after a collision between two blocks. This problem involves the conservation of momentum and energy conversion principles, including formulas for kinetic energy and spring potential energy. Suitable for students in Grades 10-12.

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