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Basic Collision/Pendulum Problem

  1. Nov 26, 2007 #1
    Ok, I need help with a basic collision problem. Please see the attached image and picture the image as a pendulum.

    You bring mass 'M' up a certain height and let it go. It collides elastically with mass 'Mew M." Mass 'm' goes up a maximum height. Find mass 'Mew M' so that mass 'm' can attain the maximum height.

    Also:
    (1) M Does not equal m.
    (2) The answer is not infinitely small, which is what i got :frown:.
     

    Attached Files:

  2. jcsd
  3. Nov 26, 2007 #2

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    Why don't you show us how you got the wrong answer? Maybe we can work it out from there.

    What things remain conserved in elastic collisions?
     
  4. Nov 26, 2007 #3
    Energy is conserved in a completely elastic collision. I used (1) energy of 'M', (2) collision of 'M' and 'm', (3), collision of 'Mew' and 'm', (4) energy of 'm'


    First see attached for description of initial height.

    So:

    (1) Energy
    Mg(L-Lcos[tex]\theta[/tex] )=(1/2)M[tex]V^{2}[/tex]

    V = [tex]\sqrt{2gL(1-cos\theta[/tex])




    (2) Find the Velocity of 'Mew' after the collision, [tex]V_{3}[/tex], with momentum equation.

    [tex]V_{3} = \frac{M(V-V_{2})}{Mew}[/tex]

    [tex]V_{2} [/tex] = Velocity of M after collision with Mew




    (3) Find the velocity of 'm' after the collision with 'Mew,' [tex] V_{5}[/tex] with momentum equations.

    [tex]V_{5} = \frac{M(V-V_{2}) - MewV_{4}}{m} [/tex]

    [tex] V_{4} [/tex] = the velocity of 'm' after the collision with 'Mew'





    (4) Use energy equation to find Hmax.

    [tex] mgHmax = \frac{1}{2} m V_{5}^{2} [/tex]




    Am I on the right track? Or am I missing something?
     

    Attached Files:

  5. Nov 27, 2007 #4

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    You need not worry about the height. The only thing of concern is the KE and V of the mass M at the moment of collision. The values of the KE etc is immaterial, because you have to find mew-m in terms of M and m.

    When a mass M collides with another mass, say M', what is the value of M' such that the whole KE of M is transferred to M' ? Think on these lines.
     
  6. Nov 27, 2007 #5
    I got it?

    When a mass M collides with another mass, say M', what is the value of M' such that the whole KE of M is transferred to M' ? Think on these lines.

    M transfers all its kinetic energy to M' only if M' is the same mass as M!! So the answer is Mew-M's mass must be equivalent to the M's mass. Or will 'm' go further if Mew-M was equal to m and not M? M does not equal m.
     
  7. Nov 27, 2007 #6

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    Not mew_m' mass but the combined mass of mew_m and m!

    Let's assume M comes to rest and gives up all the KE to mew-m and m. Then mew_m and m must add up to M.

    You may ask whether that'll ensure the mass m receiving the max energy. You see, not only the KE but the momentum MV of the mass M is also being transferred to mew_m+m. Both conditions combined make m recv the max KE, and so swing up the highest.

    (The delay in my first reply was due to a messy piece of algebra I had to before I was satisfied that this was indeed the case.)
     
    Last edited: Nov 27, 2007
  8. Nov 27, 2007 #7
    I understand your answer conceptually, but I'm not sure how to solve it algebraically. Should I start with a collision between M and Mew (which will yield the same velocity) and then Mew with m?


    Also, initially I was thinking that m must be:

    m = [tex]\frac{1}{2}[/tex] M + [tex]\frac{1}{2}[/tex] Mew

    Ah, its so open ended without math ! :cry:
     
  9. Nov 27, 2007 #8

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    Let MV = p and (1/2)MV^2 = k, for easier wrtining. I'll call mew_m as m1 and its speed as v1 etc.

    m1v1^2 + mv^2 = 2k and m1v1 + mv = p. k and p are constants.

    Eliminate v1 from the two eqns and maximise v. We want v to be max because m is a constant in the problem, so (1/2)mv^2 will be max.

    That'll give you the reqd condition.
     
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