A 14,000-kg boxcar is coasting at 1.50 m/s along a horizontal track when it suddenly hits and couples with a stationary 10,000-kg boxcar. What is the speed of the cars just after the collision? m/s

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Chapter6: Momentum And Collisions
Section: Chapter Questions
Problem 29P: a man of mass m1 = 70.0 kg is skating at v1 = 8.00 m/s behind his wife of mass m2 = 50.0 kg, who is...
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### Physics Problem: Conservation of Momentum

**Problem Statement:**

A 14,000-kg boxcar is coasting at 1.50 m/s along a horizontal track when it suddenly hits and couples with a stationary 10,000-kg boxcar.

**Question:**

What is the speed of the cars just after the collision?

**Answer Input:** 
```
[______________] m/s
```

---

**Explanation for Students:**

This problem involves the concept of conservation of momentum. When two objects collide and stick together, their combined momentum after the collision must equal the total momentum before the collision, assuming no external forces act on the system.

In this scenario:
- The first boxcar has a mass, \( m_1 = 14,000 \) kg and an initial velocity, \( v_1 = 1.50 \) m/s.
- The second boxcar has a mass, \( m_2 = 10,000 \) kg and is stationary, so its initial velocity, \( v_2 = 0 \) m/s.

We can use the conservation of momentum formula:
\[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \]

Where:
- \( v_f \) is the final velocity of both boxcars after the collision.

Plugging in the values:
\[ 14,000 \times 1.50 + 10,000 \times 0 = (14,000 + 10,000) v_f \]
\[ 21,000 = 24,000 v_f \]
\[ v_f = \frac{21,000}{24,000} \]
\[ v_f = 0.875 \text{ m/s} \]

Thus, the speed of the cars just after the collision is 0.875 m/s.
Transcribed Image Text:### Physics Problem: Conservation of Momentum **Problem Statement:** A 14,000-kg boxcar is coasting at 1.50 m/s along a horizontal track when it suddenly hits and couples with a stationary 10,000-kg boxcar. **Question:** What is the speed of the cars just after the collision? **Answer Input:** ``` [______________] m/s ``` --- **Explanation for Students:** This problem involves the concept of conservation of momentum. When two objects collide and stick together, their combined momentum after the collision must equal the total momentum before the collision, assuming no external forces act on the system. In this scenario: - The first boxcar has a mass, \( m_1 = 14,000 \) kg and an initial velocity, \( v_1 = 1.50 \) m/s. - The second boxcar has a mass, \( m_2 = 10,000 \) kg and is stationary, so its initial velocity, \( v_2 = 0 \) m/s. We can use the conservation of momentum formula: \[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \] Where: - \( v_f \) is the final velocity of both boxcars after the collision. Plugging in the values: \[ 14,000 \times 1.50 + 10,000 \times 0 = (14,000 + 10,000) v_f \] \[ 21,000 = 24,000 v_f \] \[ v_f = \frac{21,000}{24,000} \] \[ v_f = 0.875 \text{ m/s} \] Thus, the speed of the cars just after the collision is 0.875 m/s.
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