Use the worked example above to help you solve this problem. A car with mass 1.55 x 10³ kg traveling east at a speed of 20.8 m/s collides at an intersection with a 2.55 x 10³ kg van traveling north at a speed of 19.2 m/s, as shown in the figure. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that friction between the vehicles and the road can be neglected. m/s ° counterclockwise from the +x-axis magnitude direction

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
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**Goal**  
Analyze a two-dimensional inelastic collision.

**Problem**  
A car with mass 1.50 x 10³ kg traveling east at a speed of 25.0 m/s collides at an intersection with a 2.50 x 10³ kg van traveling north at a speed of 20.0 m/s, as shown in the figure. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that friction between the vehicles and the road can be neglected.

**Strategy**  
Use conservation of momentum in two dimensions. (Kinetic energy is not conserved.) Choose coordinates as in the figure. Before the collision, the only object having momentum in the x-direction is the car, while the van carries all the momentum in the y-direction. After the totally inelastic collision, both vehicles move together at some common speed vₓ and angle θ. Solve for these two unknowns, using the two components of the conservation of momentum equation.

---

**Solution**

1. **Find the x-components of the initial and final total momenta.**

   \( \Sigma p_{xi} = m_{\text{car}}v_{\text{car}} = (1.50 \times 10^3 \, \text{kg})(25.0 \, \text{m/s}) = 3.75 \times 10^4 \, \text{kg} \cdot \text{m/s} \)

   \( \Sigma p_{xf} = (m_{\text{car}} + m_{\text{van}})v_{f} \cos \theta = (4.00 \times 10^3 \, \text{kg})v_{f} \cos \theta \)

   Set the initial x-momentum equal to the final x-momentum.

   (1) \( 3.75 \times 10^4 \, \text{kg} \cdot \text{m/s} = (4.00 \times 10^3 \, \text{kg})v_{f} \cos \theta \)

2. **Find the y-components of the initial and final total momenta.**

   \( \Sigma p_{iy} = m_{\text{van}}v_{\text{van}} = (2.50 \times
Transcribed Image Text:**Goal** Analyze a two-dimensional inelastic collision. **Problem** A car with mass 1.50 x 10³ kg traveling east at a speed of 25.0 m/s collides at an intersection with a 2.50 x 10³ kg van traveling north at a speed of 20.0 m/s, as shown in the figure. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that friction between the vehicles and the road can be neglected. **Strategy** Use conservation of momentum in two dimensions. (Kinetic energy is not conserved.) Choose coordinates as in the figure. Before the collision, the only object having momentum in the x-direction is the car, while the van carries all the momentum in the y-direction. After the totally inelastic collision, both vehicles move together at some common speed vₓ and angle θ. Solve for these two unknowns, using the two components of the conservation of momentum equation. --- **Solution** 1. **Find the x-components of the initial and final total momenta.** \( \Sigma p_{xi} = m_{\text{car}}v_{\text{car}} = (1.50 \times 10^3 \, \text{kg})(25.0 \, \text{m/s}) = 3.75 \times 10^4 \, \text{kg} \cdot \text{m/s} \) \( \Sigma p_{xf} = (m_{\text{car}} + m_{\text{van}})v_{f} \cos \theta = (4.00 \times 10^3 \, \text{kg})v_{f} \cos \theta \) Set the initial x-momentum equal to the final x-momentum. (1) \( 3.75 \times 10^4 \, \text{kg} \cdot \text{m/s} = (4.00 \times 10^3 \, \text{kg})v_{f} \cos \theta \) 2. **Find the y-components of the initial and final total momenta.** \( \Sigma p_{iy} = m_{\text{van}}v_{\text{van}} = (2.50 \times
**LEARN MORE**

**REMARKS**  
It's also possible to first find the x- and y-components \( v_{fx} \) and \( v_{fy} \) of the resultant velocity. The magnitude and direction of the resultant velocity can then be found with the Pythagorean theorem,  
\[ v_f = \sqrt{v_{fx}^2 + v_{fy}^2} \],  
and the inverse tangent function \(\theta = \tan^{-1}(v_{fy}/v_{fx})\). Setting up this alternate approach is a simple matter of substituting \( v_{fx} = v_f \cos \theta \) and \( v_{fy} = v_f \sin \theta \) in Equations (1) and (2).

**QUESTION**  
If the car and van had identical mass and speed, what would the resultant angle have been?  
\[ \_\_\_\_\_\_\_\_ \] °

---

**PRACTICE IT**

Use the worked example above to help you solve this problem. A car with mass \( 1.55 \times 10^3 \) kg traveling east at a speed of 20.8 m/s collides at an intersection with a \( 2.55 \times 10^3 \) kg van traveling north at a speed of 19.2 m/s, as shown in the figure. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that friction between the vehicles and the road can be neglected.

- Magnitude: \[ \_\_\_\_\_\_\_\_ \] m/s
- Direction: \[ \_\_\_\_\_\_\_\_ \] ° counterclockwise from the +x-axis

---

**EXERCISE**    

A 3.22 kg object initially moving in the positive x-direction with a velocity of +4.99 m/s collides with and sticks to a 2.23 kg object initially moving in the negative y-direction with a velocity of -2.56 m/s. Find the final components of velocity of the composite object. (Indicate the direction with the sign of your answer.)

- \( v_{fx} = \_\_\_\_\_\_\_\_ \) m/s
- \( v_{fy} =
Transcribed Image Text:**LEARN MORE** **REMARKS** It's also possible to first find the x- and y-components \( v_{fx} \) and \( v_{fy} \) of the resultant velocity. The magnitude and direction of the resultant velocity can then be found with the Pythagorean theorem, \[ v_f = \sqrt{v_{fx}^2 + v_{fy}^2} \], and the inverse tangent function \(\theta = \tan^{-1}(v_{fy}/v_{fx})\). Setting up this alternate approach is a simple matter of substituting \( v_{fx} = v_f \cos \theta \) and \( v_{fy} = v_f \sin \theta \) in Equations (1) and (2). **QUESTION** If the car and van had identical mass and speed, what would the resultant angle have been? \[ \_\_\_\_\_\_\_\_ \] ° --- **PRACTICE IT** Use the worked example above to help you solve this problem. A car with mass \( 1.55 \times 10^3 \) kg traveling east at a speed of 20.8 m/s collides at an intersection with a \( 2.55 \times 10^3 \) kg van traveling north at a speed of 19.2 m/s, as shown in the figure. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that friction between the vehicles and the road can be neglected. - Magnitude: \[ \_\_\_\_\_\_\_\_ \] m/s - Direction: \[ \_\_\_\_\_\_\_\_ \] ° counterclockwise from the +x-axis --- **EXERCISE** A 3.22 kg object initially moving in the positive x-direction with a velocity of +4.99 m/s collides with and sticks to a 2.23 kg object initially moving in the negative y-direction with a velocity of -2.56 m/s. Find the final components of velocity of the composite object. (Indicate the direction with the sign of your answer.) - \( v_{fx} = \_\_\_\_\_\_\_\_ \) m/s - \( v_{fy} =
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