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Math Help - Help with physics

  1. #1
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    Help with physics

    A blue sports car (m=1200-kg) traveling north at 14 m/s is rear-ended by a 2000-kg truck traveling north at 25 m/s. If the truck and car locks bumpers and stick together, what is their speed immediately after the collision?


    a. What is the momentum of the car BEFORE the collison
    b. What is the momentum of the truck BEFORE the collision?
    c. What is the total momentum BEFORE the collision?
    d. What is the total momentum AFTER the collison?
    e. What is the velocity of the cars after the collison?


    Need help with the equations used in this
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  2. #2
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    Re: Help with physics

    Quote Originally Posted by Louisana1 View Post
    A blue sports car (m=1200-kg) traveling north at 14 m/s is rear-ended by a 2000-kg truck traveling north at 25 m/s. If the truck and car locks bumpers and stick together, what is their speed immediately after the collision?


    a. What is the momentum of the car BEFORE the collison
    b. What is the momentum of the truck BEFORE the collision?
    c. What is the total momentum BEFORE the collision?
    d. What is the total momentum AFTER the collison?
    e. What is the velocity of the cars after the collison?


    Need help with the equations used in this
    In the law of conservation of momentum we believe! Therefore the momentum in our system is the before and after the crash. Define an x-axis that is positive in the northern direction. The momentum of the car before the crash is
    p = mv => p = 1200*14 = 16800 kg*m/s due north

    And the momentum for the truck before the crash is
    p = mv => p = 2000*25 = 50000 kg*m/s due north

    The total momentum in the system before the crash is therefore
    p_tot = p_car+p_truck => p_tot = 16800+50000 = 66800 kg*m/s due north

    Due the law of conservation of momentum we have this momentum after the crash as well. And we also know that the speed of the car and the truck is the same. Hence the momentum afterwards can be described as
    p_tot = (m_car+m_truck)v => 66800 = 3200*v <=> v ≈ 21 m/s due north
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    Re: Help with physics

    how did you get 3200 in the last one
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    fkf
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    Re: Help with physics

    Quote Originally Posted by Louisana1 View Post
    how did you get 3200 in the last one
    Add the mass of the truck to the mass of the car, to get the total mass of the car/truck after the crash, and hence the mass of the moving object with a velocity v which we were searching.
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    Re: Help with physics

    ok for the E
    do i divide 66800/16800
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    fkf
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    Re: Help with physics

    Quote Originally Posted by Louisana1 View Post
    ok for the E
    do i divide 66800/16800
    ?
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