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Collision involving blocks and pulley

  1. May 18, 2015 #1
    1. The problem statement, all variables and given/known data

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    2. Relevant equations


    3. The attempt at a solution

    When the particle of mass M/2 collides with the pan ,an impulsive tension acts on the (pan+particle) . Let us call this J . Tension acting in the string between the two blocks is T .

    Speed of particle before collision = u = √(gl)
    Speed of (pan+particle) after collision = v

    Applying impulse momentum theorem on (pan+particle), Mu/2 -J = Mv/2

    Applying impulse momentum theorem on block A, J = Mv

    From the above two equations we get v = √(gl)/3 .

    Is my reasoning correct ? How should I proceed ? I would be grateful if somebody could help me with the problem .
     

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  3. May 18, 2015 #2

    andrevdh

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    Since A and B is connected by a taught string shouldn't the magnitude of their acceleration be the same?
    Problem is the tension in the string between A and B also changes during the collision.
    So I am getting the impression that you need to consider the system as a whole.
    Setup looks very much like a swing/pendulum on the left.
     
    Last edited: May 18, 2015
  4. May 18, 2015 #3
    Yes .
    So , how should I proceed ?
     
  5. May 18, 2015 #4
    What would you expect to happen if you dropped a mass on the edge of a pan that is connected by a string to the ceiling?
     
  6. May 18, 2015 #5

    TSny

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    I think you are correct so far. Draw a free body diagram for each mass M. Apply 2nd law to the vertical direction for each mass.

    The vertical acceleration of the mass M on the left will consist of a superposition of two parts: one part from the change in length of the left portion of the string and one part due to the fact that the sideways motion makes the mass move in circular motion.
     
  7. May 18, 2015 #6
    Hello TSny ,

    Thanks for replying .

    Assuming tension T exists in the string connecting A and B after the collision and A moves upwards as well as towards left and B downwards .

    For A , ##T - Mg = M(a+\frac{v^2}{l})## ,

    For B , ##Mg-T = Ma##

    Does it make sense ?
     
  8. May 18, 2015 #7

    TSny

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    Yes. You are letting "##a##" be the upward acceleration of A due to the changing length (and therefore the downward acceleration of B). Looks good.
     
  9. May 18, 2015 #8

    TSny

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    Do you think that B will move downward?
     
  10. May 18, 2015 #9
    But this gives ## M(2a+\frac{v^2}{l}) = 0 ## ? Does this mean , my assumption that A going upwards was wrong . Instead B moves upwards with acceleration g/18 .

    Is that so ?
     
  11. May 18, 2015 #10

    TSny

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    I think so. I also get that B accelerates upward at g/18.
     
  12. May 18, 2015 #11

    TSny

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    Here’s another way to look at it. Treat masses A and B as one system. The gravity force on each side cancels. So, the net force accelerating the system is just the “centrifugal” force due to the swinging of A.

    So, for the system: ##F = ma## gives ##Mv^2/l = (2M)a##. So ##a = v^2/(2l)## where ##v^2 = gl/9##.
     
  13. May 18, 2015 #12
    The answer given is option b) i.e g/9 :rolleyes:
     
  14. May 18, 2015 #13

    TSny

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    Good. We must still have something to learn! I don't see a mistake, so we can hope that someone will set us straight.
     
  15. May 18, 2015 #14
    How does gravity cancels if we treat A and B as one system ? It is acting downwards on both the blocks .
     
  16. May 18, 2015 #15

    TSny

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    Weight of A tries to make the system move counterclockwise while weight of B acts clockwise.
     
  17. May 18, 2015 #16
    We get g/9 as answer if we consider only the centripetal acceleration of block A .

    For block A , ##T - Mg = M\frac{v^2}{l}## . So, T = (10/9)Mg

    Now for block B , acceleration = net force /mass . So acc = (T - Mg) / M = g/9 .

    But the problem I see is that if this is the case , both the blocks then accelerate upwards with g/9 . Doesn't look convincing .

    I saw the solution given , and this is how the option b) is obtained .
     
  18. May 18, 2015 #17

    TSny

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    I agree. This solution appears to neglect the downward acceleration of A due to the changing length of the string on the left.
     
  19. May 18, 2015 #18
    How does leftwards impulse on block A gives it a downwards acceleration ? I understand that it makes A move in circular path about the top pulley , but how does it provide downward acceleration to A ?

    Downwards acceleration of block A means that string length of left string increases in vertical direction and that of right string decreases in vertical direction. Isn't that so ?

    I would like to understand your reasoning .
     
    Last edited: May 18, 2015
  20. May 18, 2015 #19

    TSny

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    The string on the left side is accelerating downward, but mass A has a total acceleration upward due to the upward centripetal acceleration. For mass A, the net upward acceleration is

    ##a_{net, A} = v^2_A/l - a##

    where ##a## is the downward contribution to the acceleration of A due to the lengthening of the string on the left. Note, ##a = a_{B}## where ##a_B## is the upward acceleration of B.

    2nd law for mass A: ##T-Mg = M(v_A^2/l-a_B)##

    For mass B: ##T-Mg = Ma_B##.

    These are the essentially the same as you already set up.
     
    Last edited: May 18, 2015
  21. May 18, 2015 #20
    Sorry...but some confusion has crept in :confused:.

    So,basically both A and B have a net acceleration g/18 upwards . Right ?

    How ? I can't visualize it :oldeek: .

    I still do not see how leftwards movement of A causes vertically lengthening of the left string . I can see horizontal lengthening of left string causing an overall lengthening of left string .Please help me understand how A accelerates downwards i.e how string length of left string increases in vertical direction.
     
    Last edited: May 18, 2015
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