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Help on Momentum problem

  1. May 28, 2015 #1
    1. The problem statement, all variables and given/known data
    A 3 kg body (mass1) moving at 4 m/s makes an elastic collision with a stationary body(mass2) of mass 2 kg. Find the velocity of each body after the collision

    2. Relevant equations
    w=delta e

    3. The attempt at a solution

    so because it is elastic collision, it means that kinetic energy is conserved... we can do the following:
    [itex] W=\Delta E [/itex]
    [itex] 0= \Delta KE_1 + \Delta KE_2 [/itex]
    [itex] 0= \frac{1}{2}m_1v_{f1}^2-\frac{1}{2}m_1v_{i1}^2+\frac{1}{2}m_2v_{2f}^2-\frac{1}{2}m_2v_{2i}^2 [/itex]
    [itex] v_{f1} = \sqrt{\frac{m_1v_{i1}^2-m_2v_{2f}^2}{m_1}} [/itex]
    ugh latex wont work:
    now we can use momentum conservation to say

    [itex]m_1v_{i1}=-m_1v_{f1}+m_2v_{f2} [/itex]
    we know vf1 from the energy part so we have the following:
    m_1v_{i1}=m_1\sqrt{\frac{m_1v_{i1}^2-m_2v_{f2}^2}{m_1}+m_2v_{f2} [/itex]

    so i have one unknown ([itex] v_{f2} [/itex]) and one equation, but that seems awfully horrible to eliminate the final velcity of mass 2. am i even doing it correctly?
    Last edited: May 28, 2015
  2. jcsd
  3. May 28, 2015 #2


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    Your equation should be m_1V_1=m_1V_1'+ m_2V_2

    Think about it, why would the second m_1V_1 be negative?

    Also I think you are confused on this equation:
    the second mass initially has a velocity of zero.
    Last edited: May 29, 2015
  4. May 28, 2015 #3
    i thought it would be negative becasue it is going in the other direction...
    i know v2i is 0.... thats why i just took it out by the next step (its not in the answer inside my square root).
  5. May 28, 2015 #4


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    You don't mean that, I hope.
    I assume your point is that it is better to keep a constant definition of which direction is positive. I agree.
    Last edited: May 28, 2015
  6. May 29, 2015 #5


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    I meant that the second V_1 should be (V_1)' (or V_1 final)

    I always thought that both would continue to go in the same direction after collision? Since the first one has no momentum, they will both share a momentum in the same direction won't they?
  7. May 29, 2015 #6
    actually, yes youre right. because the mass of the object being hit is smaller, they will go in the same direction. i should have changed that to a plus sign.
  8. May 29, 2015 #7
    A simpler notation may help.

    One can call the final velociies u and w and substitute for the masses right away.

    Then conservation of momentum and conservation of KE will give two equations in u and w which can be solved.
  9. May 29, 2015 #8


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    Whether they will continue in the same direction depends on the relative masses. In this case, as toesockshoe says, they will
  10. May 29, 2015 #9
    Make terms by the same masses and divide by part. You make a 1st class equation system.
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