You are at a train yard observing trains (because why not). You see a train car (let’s call it car 1) moving to the right ( x direction) tow

Question

You are at a train yard observing trains (because why not). You see a train car (let’s call it car 1) moving to the right ( x direction) towards a stationary train car (let’s call this one car 2). Car 1 has an initial velocity of 15.0 m/s. A helpful train employee tells you that Car 1 also has a mass of 1,825 kg and Car 2 has a mass of 2,645 kg. Car 1 gently collides with Car 2, allowing them to connect. After the collision the two train cars stay connected. You can assume that there is no friction in the system. If you have never see train cars connect, you can watch the first 25ish seconds of this video to see two train cars couple. However, these cars have friction, so they stop – unlike our problem. What is the Final Velocity of the system consisting of Car 1 and Car 2

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Sapo 3 years 2021-09-03T03:52:27+00:00 1 Answers 13 views 0

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  1. thienthanh
    0
    2021-09-03T03:54:13+00:00
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    Answer:

    6.12 m/s

    Explanation:

    Using the law of conservation of momentum

    momentum before collision = momentum after collision

    m₁v₁ + m₂v₂ = (m₁ + m₂)V    (since the train cars become attached to each other) where m₁ = mass of car 1 = 1,825 kg, m₂ = mass of car 2 = 2,645 kg, v₁ = initial velocity of car 1 = + 15.0 m/s (positive since it is moving in the positive x direction), v₂ = initial velocity of car 2 = 0 m/s (since it is initially stationary) and V = velocity of both cars after collision,

    So, m₁v₁ + m₂v₂ = (m₁ + m₂)V  

    m₁v₁ + m₂(0 m/s) = (m₁ + m₂)V  

    m₁v₁ + 0 = (m₁ + m₂)V  

    V = m₁v₁/(m₁ + m₂)

    substituting the values of the other variables into the equation, we have

    V = 1,825 kg × 15.0 m/s/(1,825 kg + 2,645 kg)

    V = 27375 kgm/s/ 4470kg

    V = 6.124 m/s

    V ≅ 6.12 m/s

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