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Conservation of momentum of two particles

  1. Jul 6, 2012 #1
    Two particles A and B of mass 4 kg and 2 kg respectively are connected by a light inextensible string. The particles are at rest on a smooth horizontal plane with the string slack. Particle A is projected directly away from B with speed um s−1. When the string goes taut the impulse transmitted through the string has magnitude 6 N s. Find

    a the common speed of the particles just after the string goes taut,

    b the value of u.


    Working:

    My diagram:
    2lazfbb.png

    Taking --> as positive:

    For A:

    6 = 4v -(4*-u)
    6 = 4v + 4u
    (6-4v)/4 = u

    Conversation of momentum:

    (-u*4) + (2*0) = (4*v) + (2*-v)
    -4u = 4v -2v
    -4u = 2v
    Subbing U from before into it I get
    -4 + 4v = 2v
    v = 3ms^-1

    this is correct

    for part b) I tried to use that u = (6-4v)/4 and subbing v into it, but I got that answer of -1.5 which is incorrect as the answer should be 4.5
     
  2. jcsd
  3. Jul 6, 2012 #2

    Doc Al

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    You're getting your signs mixed up. Note that if you take to the right as positive, then v must be negative.

    Instead, just use u and v to represent the magnitudes of the velocities. And realize that the impulse reduces the speed of A, thus u > v.
     
  4. Jul 6, 2012 #3
    Both velocities must be negative? If both velocity are in the same direction then I seem to get the right answer, why is that?
     
  5. Jul 6, 2012 #4

    Doc Al

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    The velocities are in the same direction. (To the left.)
     
  6. Jul 6, 2012 #5
    Conservation of momentum = linear shift invariance.
     
  7. Jul 6, 2012 #6
    how do you know?
     
  8. Jul 6, 2012 #7

    TSny

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    synkk,

    You assumed that mass A ends up going to the right. This can't be true. If mass A ends up going to the right with the same speed as B ends up going to the left, then the final total momentum of the system would be toward the right (since A has the greater mass). But the initial total momentum of the system is toward the left. So, total momentum would not be conserved.

    You will need to assume that both A and B end up moving to the left with the same speed.

    (You can easily get the final speed of B by applying the impulse-momentum theorem to B.)
     
  9. Jul 6, 2012 #8

    Doc Al

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    Since A starts out going to the left, the total momentum of the system will always be to the left. In particular, the common velocity when the string goes taut will be to the left.
     
  10. Jul 6, 2012 #9

    TSny

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    synkk,

    The problem only stated that the two blocks would have the same final speed. So, it might not be obvious at first which way block A would end up moving. But, hopefully you now see why block A must end up moving to the left in this problem.

    If the mass of block A had been 2 kg while that of B had been 4 kg, then you would have been correct in assuming block A ends up moving to the right.
     
  11. Jul 6, 2012 #10
    If an impulse is applied to an object initially at rest, what is the final velocity of the object?
    Since the string is taut, both object move at same rate.
    Now you have the final momentum and you can find the initial momentum.
     
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