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Linear Momentum and Conservation of Energy Question

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data
    a steel ball of m mass is fastened to light cord of lenght L and released when the cord is horizontal. At the bottom if its path the ball strikes a hard plastic block of mass M=4m, initially at rest on a frictionless surface. The collision is elastic. Find the speed of the block immediatey after the collision.


    2. Relevant equations
    KE+U+W=KEF+UF
    m1v1+m2v2=m1v3+m2v4


    3. The attempt at a solution
    I get the as speed (v1+4v2-v3)/4 but the book gets 2/5√2gL

    If anyone can help it would be much appreciated.
     
    Last edited: Oct 2, 2009
  2. jcsd
  3. Oct 2, 2009 #2

    Delphi51

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    I'm getting 2/3*sqrt(2gL). Closer than you, I think!
    The first step is the falling ball. Energy will be conserved so you can say
    PE at top = KE at bottom
    You'll find the velocity of the ball before it hits is sqrt(2gL), which is a good start.

    Then do your collision. Both the ball and the block will be moving after the collision so you'll need a v1 and a v2 - two unknowns. You were also given that it is an elastic collision, so kinetic energy is conserved. That gives you two equations with the two unknowns. After subbing one into the other you get a quadratic equation but there is no constant term so it can be common factored.
     
  4. Oct 2, 2009 #3
    So if i plug in sqrt(2gL)for v1 i get m1*sqrt(2gL)=m1v2+4m1v3,
    then factor out m1 and that gives you sqrt(2gL)=v2+4v3,
    then solve for v3, (sqrt(2gL)-v2)/4, i still dont get 2/5√2gL
     
  5. Oct 3, 2009 #4

    Delphi51

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    Agree so far. You have two unknowns so you have to have another equation. Use conservation of kinetic energy.
     
  6. Oct 3, 2009 #5
    So how would i go about setting that up would it be .5m(sqrt(2gL))^2=.5mv2^2+.54mv3^2
     
  7. Oct 3, 2009 #6

    Delphi51

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    Yes. Extra 4 in the last term - should be .5, not .54.
    Solve your momentum equation for v2 and sub it into this one to get an equation with only one unknown, v3. Solve the quadratic equation to find it.

    I'm signing off for the night. Good luck!
     
  8. Oct 3, 2009 #7
    I think you need the 4 because its four times the mass
     
  9. Oct 3, 2009 #8
    So i plug v2 value into the m1v1=m1v2+4m1v3 and solve for v3,

    now i get ((2sqrt(2gL))-(sqrt6mgl))/8

    i still think im off but am i getting closer
     
  10. Oct 3, 2009 #9

    Delphi51

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    You have two equations:
    [1] sqrt(2gL)=v2+4v3 from momentum conservation
    [2] .5m(sqrt(2gL))^2=.5mv2^2+.5*4mv3^2 from energy cons.
    You want to find v3, so you must solve one equation for v2 and sub into the other to get an equation with only unknown v3. It is easier to solve [1] for v2.

    I don't think v3 = ((2sqrt(2gL))-(sqrt6mgl))/8 is correct; units don't match for one thing.
     
  11. Oct 3, 2009 #10
    The equation the book get is v2'=(2m1/m1+4m1)*v1 but im not sure how they got to that every time i solve it i get something different
     
  12. Oct 3, 2009 #11

    Delphi51

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    Maybe you should show your work here so I can check it.
    The question asks you to find the speed of the block, which you are calling v3, and you say in the first post that the answer is 2/5√2gL. I'm still showing 2/3 instead of 2/5.

    No need to find v2 at all. And this v2'=(2m1/m1+4m1)*v1 can't possibly be correct because the m1/m1 term has no units but is added to the 4m1 term which has units of kg.
     
  13. Oct 3, 2009 #12
    Ok so we solve for v and that gets us sqrt(2gL)

    Then the book say now we use conservation of Linear momentum and the fact that kinetic energy is conserved to derive expressions for the speeds of the ball and the block immediately after collision. Since the collision is elastic, head on, and the target object is at rest the velocity of the block after collision is v2'=(2m1/m1+4m1)*v1

    So that would get you v2'=(2m1/m1+4m1)*sqrt(2gL)= 2/5 sqrt(2gL)

    But i dont see how they got to that equation, i follow how you set it up but i dont see how the book gets to the v2' equation
     
  14. Oct 3, 2009 #13

    Delphi51

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    Okay, looks like what you call v3 is what the book is calling v2.

    You have two equations:
    [1] sqrt(2gL)=v2+4v3 from momentum conservation
    [2] .5m(sqrt(2gL))^2=.5mv2^2+.5*4mv3^2 from energy cons.
    Divide all terms by .5m to simplify: (sqrt(2gL))^2=v2^2+4v3^2
    square cancels out square root" 2gL = v2^2+4v3^2

    Solve [1] for v2: v2 = sqrt(2gL) - 4*v3
    Sub in [2]: 2gL = [sqrt(2gL) - 4*v3]^2 + 4v3^2
    Now you have an equation with only one unknown, v3 (which the anwwer book calls v2). You have to square that middle term as in
    (A+B)^2 = A^2 + 2AB + B^2. Collect like terms and see what your quadratic equation looks like. Might look more familiar if you used "x" instead of "v3".
     
  15. Oct 3, 2009 #14
    Thanks ive got it now, sorry for taking up your time but that problem has been bugging me for a few days
     
  16. Oct 3, 2009 #15

    Delphi51

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    Most welcome.
     
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