You have three identical train cars, each of mass 2,618 kg, that are coupled (attached) together. These cars are moving north at a speed of 5.6 m/s. A fourth train car, which has the same mass, is moving north at a speed of 7.6 m/s, and when it catches up with the other cars, couples to that set, so that all four are now coupled. One minute later, the four cars now hit a fifth identical car that was at rest on the tracks, causing it to couple with the rest. What is the speed of the five car train? Your Answer: units Answer

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**Physics Problem: Conservation of Momentum in Train Car Collisions**

You have three identical train cars, each of mass 2,618 kg, that are coupled (attached) together. These cars are moving north at a speed of 5.6 m/s. A fourth train car, which has the same mass, is moving north at a speed of 7.6 m/s, and when it catches up with the other cars, couples to that set, so that all four are now coupled. One minute later, the four cars now hit a fifth identical car that was at rest on the tracks, causing it to couple with the rest.

**Question:**
What is the speed of the five-car train?

**Your Answer:**
[  ______  ]   [units]

To solve this problem, use the principle of conservation of momentum.

**Momentum Before and After Coupling:**
1. **Initial Momentum:**
    - Three cars moving at 5.6 m/s:
      \( p_1 = 3 \times 2,618\, \text{kg} \times 5.6\, \text{m/s} \)
    - One car moving at 7.6 m/s:
      \( p_2 = 2,618\, \text{kg} \times 7.6\, \text{m/s} \)
2. **After the Fourth Car Couples:**
    - Combine the momentum:
      \( P_{\text{total}} = p_1 + p_2 \)
    - Total mass of four cars:
      \( M_{\text{total}} = 4 \times 2,618\, \text{kg} \)
    - Speed after coupling (\(v_4\)):
      \( v_4 = \frac{P_{\text{total}}}{M_{\text{total}}} \)
3. **After Coupling with the Fifth Car:**
    - Fifth car is initially at rest:
      \( p_3 = 0 \)
    - Total mass of five cars:
      \( M_{5} = 5 \times 2,618\, \text{kg} \)
    - Speed after coupling (\(v_5\)):
      \( v_{5} = \frac{(M_{\text{total}} \times v_4)}{M_{5}} \)

This way, by applying the conservation of momentum principle and solving step
Transcribed Image Text:**Physics Problem: Conservation of Momentum in Train Car Collisions** You have three identical train cars, each of mass 2,618 kg, that are coupled (attached) together. These cars are moving north at a speed of 5.6 m/s. A fourth train car, which has the same mass, is moving north at a speed of 7.6 m/s, and when it catches up with the other cars, couples to that set, so that all four are now coupled. One minute later, the four cars now hit a fifth identical car that was at rest on the tracks, causing it to couple with the rest. **Question:** What is the speed of the five-car train? **Your Answer:** [ ______ ] [units] To solve this problem, use the principle of conservation of momentum. **Momentum Before and After Coupling:** 1. **Initial Momentum:** - Three cars moving at 5.6 m/s: \( p_1 = 3 \times 2,618\, \text{kg} \times 5.6\, \text{m/s} \) - One car moving at 7.6 m/s: \( p_2 = 2,618\, \text{kg} \times 7.6\, \text{m/s} \) 2. **After the Fourth Car Couples:** - Combine the momentum: \( P_{\text{total}} = p_1 + p_2 \) - Total mass of four cars: \( M_{\text{total}} = 4 \times 2,618\, \text{kg} \) - Speed after coupling (\(v_4\)): \( v_4 = \frac{P_{\text{total}}}{M_{\text{total}}} \) 3. **After Coupling with the Fifth Car:** - Fifth car is initially at rest: \( p_3 = 0 \) - Total mass of five cars: \( M_{5} = 5 \times 2,618\, \text{kg} \) - Speed after coupling (\(v_5\)): \( v_{5} = \frac{(M_{\text{total}} \times v_4)}{M_{5}} \) This way, by applying the conservation of momentum principle and solving step
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