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  1. #1
    Senior Member AchiLLe..s's Avatar
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    Physics Help- Please

    Ok I am stumped on these 2 questions.

    A thread holds a 1.5-kg and a 4.50-kg cart together. After the thread is burned, a compressed spring pushes the carts apart, giving the 1.5 kg cart a speed of 32 cm/s to the left. What is the velocity of the 4.5-kg cart?

    So far my givens are

    Mass A= 1.5 kg
    Mass B= 4.50 kg
    Velocity A= 32 cm/s
    Velocity B= X

    How am I supposed to find the answer with only thoose givens using the law of conservation of momentum equations? I'm not sure what the velocity final is for either one. I'm really confused.

    As for the second question. I have the same problem

    Carmen and Judi dock a canoe. 74.0-kg Carmen moves forward at 4.0 m/s as she leaves the canoe to step onto the dock. At what speed and in what direction do the canoe and Judi move if their combined mass is 115 kg? (Use a positive number if the canoe moves away from Judi.)

    Givens:

    Mass A= 74 kg
    Velocity A= 4.0 m/s
    Mass B= 115 kg? (not sure about this one)
    Velocity B= X?

    This question I am totally clueless as well and would love if someone could point me in the right direction.

    Equations:

    Ma(Va)+Mb(Vb)= Ma(Vaf)+Mb(Vbf) Elastic equation

    Ma(Va)+Mb(Vb)= (Ma+Mb)Vf Inelastic Equation

  2. #2
    Genius FlatMaster's Avatar
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    First one. You're correct that you need to conserve momentum.

    p=mv
    What is the initial momentum?
    velocity is zero, so initial momentum is zero.

    What is the final momentum?
    Well, both carts have a velocity so momentum must not be zero right??
    Wrong!! Remember, momentum is a vector. The cart moving to the right has positive momentum, while the cart moving to the left has negative momentum.

    p initial = p final
    0 = p final
    0 = m1v1 + m2v2
    Ride or Die

  3. #3
    Zan
    Zan is offline
    Senior Member Zan's Avatar
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    second one = heavy girls

    m1v1 = m2v2

    (74)(4) = (115)v2

    v2 = 74*4/115

    pretty much like the first. just remember to take into account direction. choose which is "positive" and the other negative.

    i'm actually doing these questions in physics right now.
    -- Zan

    "Every dog needs a squeak toy."

  4. #4
    Ho-Jahm Hocam's Avatar
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    My rule for physics "When in doubt, multiply them together"

    If that doesn't work, just look at the unit's of the final answer and do some set of operations to what you're given to get the same units.
    Race-o-meter:
    Broken until next season

  5. #5
    Senior Member AchiLLe..s's Avatar
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    Thank you guys. I understand it a lot better now.

  6. #6
    riding once again jschen's Avatar
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    Quote Originally Posted by AchiLLe..s View Post
    How am I supposed to find the answer with only thoose givens using the law of conservation of momentum equations? I'm not sure what the velocity final is for either one. I'm really confused.

    [...]

    This question I am totally clueless as well and would love if someone could point me in the right direction.
    All these problems are solved EXACTLY the same way.

    Step 1: Draw a force diagram. If the solution is now clear, solve the problem.

    Step 2: If the solution is not yet clear, draw a clearer, better labeled force diagram. If the solution is now clear, solve the problem.

    Step 3: Repeat step 2 as necessary.
    If you notice this notice then you will notice that this notice is not worth noticing.

  7. #7
    Senior Member deraltekluge's Avatar
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    Conservation of momentum. If the initial momentum of a system is zero, the total final momentum must be zero, too. You know the mass and final velocity of one part of the system, so you can calculate its momentum. The other part of the system must have a momentum of the same magnitude, but in the opposite direction. You know its mass, so you can calculate its final velocity.

  8. #8
    Dude wheres my guads? skinnyone's Avatar
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    Use the conservation of momentum. Think of the 3rd law. Every action has an equal and opposite reaction. Toss in the second law. Profit.

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