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Problem regarding centre of mass and linear momentum.

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data

    A block of mass "m" is placed on a triangular block of mass "M" , which in turn is placed on a horizontal surface as shown in figure. Assuming frictionless surfaces , find the velocity of the triangular block when the smaller block reaches the bottom end.

    Figure : http://postimage.org/image/cv0qo69kd/

    2. Relevant equations

    If no external net forces act on a system :
    Conservation of linear momentum : Pf=Pi
    Also if net external forces are zero , and initially system is at rest then , by concept of centre of mass ,
    m1x1=m2x2
    m1Δx1=m2Δx2

    3. The attempt at a solution

    Now Here is what I tried so far :

    Let's take two blocks as a system , no net external force acts horizontally , centre of mass will not change its position in that direction. But how to apply conservation of momentum here ? I am totally confused. This is question too different from others I tried.

    I did 58 questions of centre of mass , and this 59th one has bashed my brain hard.

    Please help !!

    Thanks in advance... :smile:
     
  2. jcsd
  3. Jan 29, 2013 #2

    tms

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    What about gravity?
     
  4. Jan 30, 2013 #3
    Re-read. I said no net external force acts on the system in horizontal direction below "The attempt at a solution".
     
  5. Jan 30, 2013 #4
    square root of [2mgh/(M+m)]?
     
  6. Jan 30, 2013 #5

    fgb

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    I have not solved the problem, but shouldn't the answer be a function of θ? For θ=90°, for instance, we know the answer should be zero, since m would be in freefall and would not "push" the triangular block.
     
  7. Jan 30, 2013 #6

    fgb

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    [itex] v = \sqrt{\dfrac{2mgh}{M+\frac{M^2}{m\cos^2{\theta}}}} [/itex], maybe?

    Not sure my answer is correct, but here are a few hints: Use conservation of linear momentum (horizontal) to relate V (velocity of big block) to the HORIZONTAL component of v (velocity of small block).

    Use also conservation of energy. Notice that in the end you have KE for both blocks, and KE depends on the magnitude of velocity (not only the horizontal component. How can you relate v to v_horizontal?)

    UPDATE: Notice that [itex]v \rightarrow 0 [/itex] as [itex] \theta \rightarrow 90°[/itex] :)
     
  8. Jan 30, 2013 #7
    What you said is very true.
     
  9. Jan 30, 2013 #8
    Not even near.

    You're correct.

    Almost near to correct answer put thinking straightforward your answer is also incorrect. Numerator part of your answer completely matches with the correct answer , however denominator part does not. Thanks for replies anyway.

    Ok , perhaps you might have done some careless mistakes as your logic's fine. I will use your logic to see if I get the correct answer. Will get back on ya.

    Edit : In image , angle is alpha , and its intact.
     
    Last edited: Jan 30, 2013
  10. Jan 30, 2013 #9
    Hii fgb ,

    I tried all those hints , and get the same answer as yours , exactly same..

    My first attempt was as follows :

    Let the velocity of small block be v1 at bottom and the triangular block be travelling due left by v2 ,

    Applying conservation of linear momentum at bottom ,

    mv1cos(α) = Mv2
    v1=Mv2/m

    Considering conservation of mechanical energy for the system :

    mv12/2 + Mv22/2 = mgh

    Now putting v1=Mv2/m in above equation , I got the answer same as yours yet not the correct answer.

    Now here comes the second attempt :

    I realised that as small block comes down it will have two velocity vectors. One v1cos(α) and the other v2 , it will also be travelling backwards at peak point and thus , I obtained :

    mv1cos(α) = Mv2 + mv2
    v1=v2(M+m)/mcos(α)

    Then I again used conservation of mechanical energy equation. Still I got the numerator part correct but denominator part did not match.

    May be others can reply. I am clueless.
    Some hints will do...
     
    Last edited: Jan 30, 2013
  11. Jan 30, 2013 #10

    ehild

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    [itex] v = \sqrt{\dfrac{2mgh}{M+\frac{M^2}{m\cos^2{\alpha}}}}= \sqrt{\dfrac{2mgh}{M+\frac{M^2}{m}(1+\tan^2{\alpha})}}[/itex]

    ehild
     
  12. Jan 30, 2013 #11
    Hii ehild !! :smile:

    I can not understand. The answer posted by fgb is wrong and does not match with by textbook's answer , no matter how it is changed by applying trigonometry.

    Edit : Please see my working in post #9. Where did I do wrong ?
     
  13. Jan 30, 2013 #12

    fgb

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    sankalpmittal, could you post the correct answer? I know it is kind of cheating, but it might be useful to find out where we are getting it wrong :P
     
  14. Jan 30, 2013 #13
    The answer given is ,

    [{2m2ghcos2α}/{(M+m)(M+msin2α)}]1/2
     
  15. Jan 30, 2013 #14

    ehild

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    The velocity of the block with respect to the slope is parallel to the slope, the velocity in the rest frame of reference is not.

    Let be u the speed of the block with respect to the slope, travelling with velocity v2 then the velocity in the rest frame of reference is vx=v2+ucosα, the y component is vy=-usinα. You have to calculate the KE of the block from these vx and vy.

    ehild
     
  16. Jan 30, 2013 #15
    I followed your method ,

    m(vx2+vy2)/2 + Mv22/2 = mgh

    And

    mvx = Mv2

    Putting ,vx=v2+ucosα
    And vy=-usinα in above two equations , I got the answer as :

    [{2m2ghcos2α}/{M2+m2sin2α+Mm-3Mmsin2α}]1/2

    This denominator part does not match the correct answer.
     
  17. Jan 30, 2013 #16

    ehild

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  18. Jan 30, 2013 #17
     
  19. Jan 30, 2013 #18

    ehild

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    These "moving constraints" problems are really tricky. Using conservation of energy and conservation of momentum looked the easiest method to apply, and it was your method:)

    ehild
     
  20. Jan 30, 2013 #19
    You're good. You can see the different reference frames and introduce relative velocity.
     
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