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Problem on conservation of momentum and collision

  1. Sep 13, 2015 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    Three identical balls (of equal masses) are connected by light inextensible strings and kept on a smooth horizontal surface. The middle ball B is given a velocity ##v_0## at ##t=0##. Find the velocity of ball A when A collides with C.
    collison.png

    2. Relevant equations
    Impulse ##J=\Delta P##

    3. The attempt at a solution
    Since there is no net external horizontal force on the system, acceleration of Centre of Mass along ##x## is zero. Also, ##v_{cm,x}=0##.

    Velocity of A and B along the string should be equal to ##v\sin\theta## so that the string does not extend.
    c2.png

    Along ##x##, velocity of A is ##v\sin\theta\cos\theta##. If I can assume that all three balls have small radius, the final velocity of A will be equal to velocity of B which varies due to tension.

    I tried using work-energy theorem but initially Tension is impulsive and the velocity of A and B increase suddenly. (I am not sure about this but I learnt I cant use work-energy theorem when there is a sudden increase in velocity after reading one of the insight articles)

    That leaves me with conservation of momentum of the system. Momentum of ball A or C cannot be conserved due to tension. But I am finding it difficult to get the equation.
     
  2. jcsd
  3. Sep 13, 2015 #2

    Nathanael

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    I do not understand this part.

    I don't believe this is correct. You are correct that the components of VA and VB must be the same in the direction of the length of string, but there can be a nonzero component in the perpendicular direction which would contribute to the x-component of VA.

    Yes, using conservation of energy would be an assumption. If you're unsure (as am I in this case) you should solve it by other means and then check if energy was conserved.

    (Edited:)
    You could choose coordinates and try to write and solve equations for the motion in time, but there may be a simpler way.
     
    Last edited: Sep 13, 2015
  4. Sep 13, 2015 #3

    PeroK

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    You may want to forget about equations to begin with and try to figure out an important constraint that must apply when ball A collides with ball C.
     
  5. Sep 13, 2015 #4

    Titan97

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    Since length of string does not change, ##x^2+y^2=L^2##. then ##x\frac{dx}{dt}+y\frac{dy}{dt}=0## Is this the equation @PeroK ?
    When they collide, x=0.

    And,
    What I meant is, just when the ball are about to collide, the velocity of the two lower balls will be only along ##y## and will be equal to that of velocity of B.
     
  6. Sep 13, 2015 #5

    PeroK

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    Your last statement is the key. The velocity of the balls (in the y direction) when they collide must be equal to the velocity of ball B.
     
  7. Sep 14, 2015 #6

    Titan97

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    What kind of equation are you suggesting @PeroK ?
    You mean equation of trajectory of ball A and C?
     
  8. Sep 15, 2015 #7

    haruspex

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    Why?
    More likely, conservation laws.
     
  9. Sep 15, 2015 #8

    Titan97

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    It will have a component along x. But its velocity along y should be equal to velocity of ball B
     
  10. Sep 15, 2015 #9

    haruspex

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    Right.
    What conservation laws can you apply?
     
  11. Sep 15, 2015 #10

    Titan97

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    Conservation of momentum in a direction perpendicular to string?
     
  12. Sep 15, 2015 #11

    haruspex

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    Do you mean perpendicular to the direction the string takes at a particular time, or generally?
    The benefit of conservation laws is that they allow you to relate the starting state to the final state directly, without having to worry about what goes on in between.
    Yes, momentum will be conserved, but remember that momentum is a vector, so in a 2D set-up that is effectively two equations.
    What else will be conserved?
     
  13. Sep 15, 2015 #12

    Titan97

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    Momentum of the system can be conserved. But for ball A, tension is an external force. So how can momentum of ball A or C can be conserved individually?
     
  14. Sep 15, 2015 #13

    haruspex

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    I'm not suggesting their momenta are conserved individually. In terms of the unknown final velocities, write out the initial and final momenta for the total system.
     
  15. Sep 16, 2015 #14

    Titan97

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    Initially, its ##3mv_0## along y-axis.

    Finally, its ##3mv## along y-axis and 0 along x-axis?
     
  16. Sep 16, 2015 #15

    haruspex

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    how do you get that?
     
  17. Sep 17, 2015 #16

    Titan97

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    Initially, a total mass of 3m has a velocity ##v_0## upwards.
     
  18. Sep 17, 2015 #17

    Orodruin

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    This is not correct, only the middle ball has an initial velocity.
     
  19. Sep 18, 2015 #18

    Titan97

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    ##v_{cm}=\frac{mv_0}{3m}##
    So initial momentum is mv0/3 ?
     
  20. Sep 18, 2015 #19

    Orodruin

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    No, initially you have one moving mass of mass ##m## which is moving with velocity ##v_0##. What does that mean for the initial momentum?
     
  21. Sep 19, 2015 #20

    Titan97

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    Its ##mv_0##. Finally, along y-axis, its ##3mv## where ##v## is the final velocity of any ball along y-axis.
     
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