4. A 2100 kg truck is traveling east through an intersection at 2.0 m/s when it is hit simultaneously from the side and the rear. One car is a 1200 kg compact traveling north at 5.0 m/s. The other is a 1500 kg midsize traveling east at 10 m/s. The three vehicles become entangled and begin to slide as one body. What is their velocity (both magnitude and direction!) just after the collision (before they have slowed appreciably from kinetic friction)?

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### Problem Statement:

A 2100 kg truck is traveling east through an intersection at 2.0 m/s when it is hit simultaneously from the side and the rear. One car is a 1200 kg compact traveling north at 5.0 m/s. The other is a 1500 kg midsize traveling east at 10 m/s. The three vehicles become entangled and begin to slide as one body. What is their velocity (both magnitude and direction) just after the collision (before they have slowed appreciably from kinetic friction)?

### Explanation:

To solve this problem, we need to use the principles of conservation of momentum. The momentum before the collision must equal the momentum after the collision because no external forces are acting on the system.

#### Step-by-step Approach:

1. **Calculate the initial momentum** for each vehicle:
   - **Truck (2100 kg) eastward**: 
     - Momentum = Mass × Velocity = 2100 kg × 2.0 m/s = 4200 kg·m/s (east)
   - **Compact car (1200 kg) north**:
     - Momentum = Mass × Velocity = 1200 kg × 5.0 m/s = 6000 kg·m/s (north)
   - **Midsize car (1500 kg) eastward**:
     - Momentum = Mass × Velocity = 1500 kg × 10 m/s = 15000 kg·m/s (east)

2. **Combine the momenta**:
   - Total eastward momentum = 4200 kg·m/s + 15000 kg·m/s = 19200 kg·m/s
   - Total northward momentum = 6000 kg·m/s

3. **Find the resultant velocity**:
   - Total mass = 2100 kg + 1200 kg + 1500 kg = 4800 kg
   - The resultant velocity in each direction is given by the formula:  
     \[ \text{Velocity (east)} = \frac{\text{Total eastward momentum}}{\text{Total mass}} = \frac{19200}{4800} = 4.0 \text{ m/s} \]
     \[ \text{Velocity (north)} = \frac{\text{Total northward momentum}}{\text{Total mass}} = \frac{6000}{4800} \approx 1.25 \text
Transcribed Image Text:### Problem Statement: A 2100 kg truck is traveling east through an intersection at 2.0 m/s when it is hit simultaneously from the side and the rear. One car is a 1200 kg compact traveling north at 5.0 m/s. The other is a 1500 kg midsize traveling east at 10 m/s. The three vehicles become entangled and begin to slide as one body. What is their velocity (both magnitude and direction) just after the collision (before they have slowed appreciably from kinetic friction)? ### Explanation: To solve this problem, we need to use the principles of conservation of momentum. The momentum before the collision must equal the momentum after the collision because no external forces are acting on the system. #### Step-by-step Approach: 1. **Calculate the initial momentum** for each vehicle: - **Truck (2100 kg) eastward**: - Momentum = Mass × Velocity = 2100 kg × 2.0 m/s = 4200 kg·m/s (east) - **Compact car (1200 kg) north**: - Momentum = Mass × Velocity = 1200 kg × 5.0 m/s = 6000 kg·m/s (north) - **Midsize car (1500 kg) eastward**: - Momentum = Mass × Velocity = 1500 kg × 10 m/s = 15000 kg·m/s (east) 2. **Combine the momenta**: - Total eastward momentum = 4200 kg·m/s + 15000 kg·m/s = 19200 kg·m/s - Total northward momentum = 6000 kg·m/s 3. **Find the resultant velocity**: - Total mass = 2100 kg + 1200 kg + 1500 kg = 4800 kg - The resultant velocity in each direction is given by the formula: \[ \text{Velocity (east)} = \frac{\text{Total eastward momentum}}{\text{Total mass}} = \frac{19200}{4800} = 4.0 \text{ m/s} \] \[ \text{Velocity (north)} = \frac{\text{Total northward momentum}}{\text{Total mass}} = \frac{6000}{4800} \approx 1.25 \text
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