A **1350-kg** truck traveling west at **25** **m/s** collides with a **1050-kg** car traveling north at **14** **m/s.** After the collision, the vehicles do not separate. Determine the ***x*- and** ***y*-components of** the velocity of the combined vehicles immediately after the collision.

To find the \( x \)- and \( y \)-components of the velocity of the combined vehicles after the collision, we can apply the principle of conservation of momentum. Since momentum is conserved in both the \( x \)- and \( y \)-directions, we can treat them independently.

### Given Data
- Mass of the truck, \( m_{\text{truck}} = 1350 \, \text{kg} \)
- Velocity of the truck (west), \( v_{\text{truck}} = 25 \, \text{m/s} \)
- Mass of the car, \( m_{\text{car}} = 1050 \, \text{kg} \)
- Velocity of the car (north), \( v_{\text{car}} = 14 \, \text{m/s} \)

### Conservation of Momentum in the \( x \)-Direction (West)
Since the truck is moving west, we’ll assign this direction as the \( x \)-axis:
\[
p_{x_{\text{initial}}} = m_{\text{truck}} \cdot v_{\text{truck}} + m_{\text{car}} \cdot 0 = 1350 \times 25 + 0 = 33750 \, \text{kg} \cdot \text{m/s}
\]

### Conservation of Momentum in the \( y \)-Direction (North)
Since the car is moving north, we’ll assign this direction as the \( y \)-axis:
\[
p_{y_{\text{initial}}} = m_{\text{car}} \cdot v_{\text{car}} + m_{\text{truck}} \cdot 0 = 1050 \times 14 + 0 = 14700 \, \text{kg} \cdot \text{m/s}
\]

### Total Mass After Collision
Since the two vehicles stick together after the collision, their combined mass is:
\[
m_{\text{total}} = m_{\text{truck}} + m_{\text{car}} = 1350 + 1050 = 2400 \, \text{kg}
\]

### Calculating \( x \)-Component of Velocity (\( v_x \))
Using the conservation of momentum in the \( x \)-direction:
\[
v_x = \frac{p_{x_{\text{initial}}}}{m_{\text{total}}} = \frac{33750}{2400} \approx 14.06 \, \text{m/s}
\]

### Calculating \( y \)-Component of Velocity (\( v_y \))
Using the conservation of momentum in the \( y \)-direction:
\[
v_y = \frac{p_{y_{\text{initial}}}}{m_{\text{total}}} = \frac{14700}{2400} \approx 6.13 \, \text{m/s}
\]

### Final Answer
The \( x \)- and \( y \)-components of the velocity of the combined vehicles immediately after the collision are:
- \( v_x \approx 14.06 \, \text{m/s} \)
- \( v_y \approx 6.13 \, \text{m/s} \)

Would you like further clarification or details on any part of this solution?

---

Here are 5 related questions to deepen understanding:
1. How would the direction of the final velocity be calculated for the combined vehicles?
2. What if the truck was moving east instead of west; how would this change the calculation?
3. How is momentum conserved in inelastic collisions, even though kinetic energy is not?
4. How would the problem change if the vehicles did not stick together after the collision?
5. What if the car had an initial velocity component in the \( x \)-direction?

**Tip:** Always break down momentum problems into \( x \)- and \( y \)-components for easier calculations, especially when dealing with right-angle impacts.

A 1350-kg truck traveling west at 25 m/s collides with a 1050-kg car traveling north at 14 m/s. After the collision, the vehicles do not separate. Determine the x- and y-components of the velocity of the combined vehicles immediately after the collision.

Learn how to solve a momentum conservation problem involving a 1350-kg truck and a 1050-kg car. The vehicles collide in an inelastic collision, and we calculate the x- and y-components of their combined velocity immediately after the impact. Suitable for students in Grades 11-12 studying physics and momentum.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation