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What is their final velocity?

  1. Dec 16, 2004 #1

    km

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    Need help determining the following:

    A 50kg object moving at 20 m/s collides with a 10-kg object moving at 5 m/s. They stick together and move off together. What is their final velocity?

    I know how to determine this if one object was stationary, but in this problem, both are moving. Help! 6th grade science problem
     
  2. jcsd
  3. Dec 16, 2004 #2
    Hint: Conservation of Momentum. New object has mass of 60kg. If they are moving along the same axis - the calculation would be really simple
     
  4. Dec 16, 2004 #3

    km

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    Still need to know what the final velocity is and what formula to use to get the answer.

    km
     
  5. Dec 16, 2004 #4
    Does your textbook tell you about anything in regards to conservation of momentum? Take a guess as to what the formula is....

    I'm trying to guide you through it, and not just give you the answer...
     
  6. Dec 16, 2004 #5
    Use the same principle as when one object is initially stationary, but apply a more generalized formula:

    [tex]\vec{p}_{initial}=\vec{p}_{final}[/tex]
    [tex](m_{1}\vec{v}_{1})_{initial}+(m_{2}\vec{v}_{2})_{initial}=(m_{1}\vec{v}_{1})_{final}+(m_{2}\vec{v}_{2})_{final}[/tex]

    Can you go from there?
     
  7. Dec 16, 2004 #6

    km

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    Never mine I think I got it.

    Answer is 10m/s. I hope.

    Thanks anyway.

    km
     
  8. Dec 16, 2004 #7
    That is incorrect. Can you show us your work?
     
  9. Dec 16, 2004 #8

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    First I determined the total momentum:
    (5 kg x 20 m/s) + (10kg x 5 m/s) = 150 kg.ms

    then determined velocity:
    150kg.m/s = (5kg + 10kg) x velocity
    10 m/s = velocity

    this is what is in my textbook but there were no examples of both objects moving
     
  10. Dec 16, 2004 #9
    The original question you posted says one mass is 50 kg. If it is 5 kg, as the latest version of the problem states above, you are correct, assuming both objects are initially travelling in the same direction.
     
  11. Dec 16, 2004 #10

    km

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    OOPS! My mistake. Yes, the first object is 5-kg. Sorry for the mix-up on original posting. Thank you for letting me know this was correct!!!
     
  12. Dec 16, 2004 #11

    km

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    Since you mentioned Conservation of momentum earlier, I also have another question I am working on that I am having trouble with as my textbook does not talk about rocketships!

    The question is: How do you explain conservation of momentum when a rocket ship takes off?

    What I have found on the internet so far says stuff about the the rocket ship pushing on the fuel and the fuel pushing on the rocket ship but I don't understand!
     
  13. Dec 17, 2004 #12
    Momentum is conserved when we take the movement of a rocket ship with no fricitonal forces, etc.
    [tex]\vec{p}_{initial}=\vec{p}_{final}[/tex]
    [tex](m_{r}\vec{v}_{r})_{initial}+(m_{f}\vec{v}_{f})_{initial}=(m_{r}\vec{v}_{r})_{final}+(m_{f}\vec{v}_{f})_{final}[/tex]
    where r is rocket (w/o fuel) and f is fuel. Since the rocket and fuel are initially travelling at the same velocity (before the engine is engaged), we can simplify this to:
    [tex](m_{(r+f)}\vec{v}_{(r+f)})_{initial}=(m_{r}\vec{v}_{r})_{final}+(m_{f}\vec{v}_{f})_{final}[/tex]
    Keep in mind that velocity and momentum are vectors here, so we need to consider direction. If we are speaking of linear displacement, we can simply define opposite directions as positive and negative; in that case, watch your signs!
     
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