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Using the law of conservation of kinetic energy

  1. Dec 26, 2004 #1
    HI,
    I am wondering how to solve this physics question:

    -We have an object A and we have an object B
    -We place these objects at a distance from each other.
    -Object A has mass m.
    -Object B has mass 3m.
    -Object A moves toward object B with velocity v.
    -Object B moves toward object A with velocity -v.

    Q; What is the velocity of each object after impact?
    --------------------------------------------------------
    Can you tell me what I am doing wrong here: (I don't know where I have made a mistake)

    (v1) = velocity of Object A after impact.
    (v2) = velocity of Object B after impact.

    Total momentum = mv-3mv= -2mv = m(v1) + 3m(v2)

    Dividing both sides by m we find that:
    -2v = (v1) + 3(v2)

    and solving for v1 we find,

    (v1) = -2v-3(v2) (1)

    Using the law of conservation of kinetic energy we also get:

    (.5)mv^2 + (.5)(3m)v^2 = (.5)m(v1)^2 + (.5)(3m)(v2)^2

    Now divide by .5 and then by m to come up with:

    4v^2 = (v1)^2 + 3(v2)^2 (2)

    Inserting v1 from (1) into (2):

    4v^2 = (-2v-3(v2))^2 + 3(v2)^2
    4v^2 = 4v^2 + 12v(v2) + 9(v2)^2 + 3(v2)^2
    -12v(v2) = 12 (v2)^2

    and finally (dividing by (v2) and 12),
    (v2)=-v

    plugging (v2)=-v into (2) to come up with:
    (v1)= v

    It doesn't even make sense or look right and I have been trying to determine where I have made a mistake for days! :confused: :confused:
    Thanks for any help...
     
  2. jcsd
  3. Dec 26, 2004 #2

    krab

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    Science Advisor

    You only have one of the solutions, namely the one where they fail to interact. The other solution is v2=0, v1=-2v.
     
  4. Dec 26, 2004 #3
    Wow, you really caught my mistake!
    I understand now, instead of dividing by (v2) I should have brought -12v(v2) to the other side and factored to find (v2). Thanks a bunch.
     
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