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Mechanics of a block of mass homework

  1. Mar 5, 2005 #1
    A block of mass m = 2.30 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 7.10 kg which is at rest on a horizontal surface, Fig. 7-41. (Assume a smooth transition at the bottom of the incline, an elastic collision, and ignore friction.)

    :confused: Please help to solve this problem.
     
  2. jcsd
  3. Mar 5, 2005 #2

    dextercioby

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    What's the question...?You didn't post the question..

    Daniel.
     
  4. Mar 5, 2005 #3
    Sorry. Determine the speed of the smaller mass and the larger mass after collision?
     
  5. Mar 5, 2005 #4

    dextercioby

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    Okay then.The advice is simple:use the conservation of KE and momentum (in vector form).Then u'll have to solve a simple system.

    Daniel.
     
  6. Mar 5, 2005 #5
    I did. I end up with two unknown values for a simultaneuos equation. But I keep getting the wrong answer. Do you have an answer to this question?
     
  7. Mar 5, 2005 #6

    dextercioby

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    Nope.I'm just trying to guide through the correct solving.It's your problem.

    So post the equations & the reasoning for reaching them in the first place;

    Daniel.
     
  8. Mar 5, 2005 #7
    Ok. I use the Gravitation potential Energy lost=Kinetic energy lost rule to determine the speed of the smaller mass before the collision. I get 8.40m/s. Is this method acceptable to work out the initial speed of the smaller mass?
     
  9. Mar 5, 2005 #8

    dextercioby

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    Yes,the initial value for the speed of the small mass is 8.40m/s...

    Daniel.
     
  10. Mar 5, 2005 #9
    Good. Now i can build a simultaneous equation. Equation one deals witht the momentum of the system. Equation two deals with the kinetic energy of the system.

    2.30x8.40 + 0 = 2.30v + 7.10V ..........(1)
    0.5x2.30x8.40^2 + 0 = 0.5x2.30xv^2 + 7.10V^2..........(2)
    Is this correct so far?
     
  11. Mar 5, 2005 #10
    sorry i have to correct equation (2)
    0.5x2.30x8.40 + 0 = 0.5x2.30xv^2 + 0.5x7.1xV^2
     
  12. Mar 5, 2005 #11
    is this correct so far then?
     
  13. Mar 5, 2005 #12

    dextercioby

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    Assuming that the big mass is not exactly at the botton of the incline,then yes.

    Daniel.
     
  14. Mar 5, 2005 #13
    Well from rearranging equation (2) to give V in terms of v, I end up with the answer
    v=8.00m/s.
    This is an electronic Physics homework that my school sets every week. After submitting it, the program tells me that my answer is wrong.
     
  15. Mar 5, 2005 #14

    dextercioby

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    Well,as u see,since a smaller object is collided with a bigger one,it will bounce back,which means that the velocity vector changes the sense...Which means that the final aswer should e negative...

    Daniel.

    P.S.Therefore,take "-v" in the momentum conservation law.
     
  16. Mar 5, 2005 #15
    But the question doesnt talk about velocity. The question asks for speed, which is not a vecor quantity. So just 8.00m/s should be fine. Maybe I am giving my answers to the wrong numbers of decimal places or significant figures?
    Is my my speed correct though?
     
  17. Mar 5, 2005 #16

    dextercioby

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    Choosing "-v" in the component form of the momentum conservation eq.you'll end up with v~5.33ms^{-1}...

    Daniel.
     
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