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Homework Help: Conservation of Linear Momentum?

  1. Nov 16, 2005 #1
    Particle A & B are held together with a compressed spring between them. When they are released the spring pushes them apart and they fly off in opposite directions free of the spring. The mass of A is twice that of B, and the energy stored in the spring was 60J. Once the transfer is complete, what are the kinetic energies of particle A & B

    I have no clue where to start on this one.

    Can Anyone Point Me In The Right Direction?

  2. jcsd
  3. Nov 16, 2005 #2
    The title of your post gives it away.
    There are two conditions that must be met:
    That the momentum of particle A is equal and opposite to the momentum of particle B. (conservation of momentum)
    That the combined kinetic energy of particle A & B is equal to 60J (conservation of energy)

    Write out this two conditions in terms of mass and velocity and you should be able to solve.
  4. Nov 16, 2005 #3
    Just wondering you know if the answer is in symbolic form or do you have numbers such as the weight of b?
  5. Nov 16, 2005 #4
    There are no weights given for the two particles
  6. Nov 16, 2005 #5
    Im still not too sure what to do.

    How do I get the values of Ke
  7. Nov 16, 2005 #6
    Would I use:

    [tex]W=\Delta {E}_m_e_c+\Delta {E}_t_h_e[/tex]

    Also I dont have a [tex]\vec{v}[/tex]
    Last edited: Nov 16, 2005
  8. Nov 16, 2005 #7
    ?can Anyone Help?
  9. Nov 17, 2005 #8
    Sorry, been away. Ok, advanced reply doesn't seem to be working so hopefully my latex comes out ok.
    We'll go through the two conditions (conservation of momentum/energy) seperately then combine them.

    Conservation of momentum:
    The momentum of the system doesn't change. Assuming the system is stationary to begin with.


    [tex]m_a*v_a + -m_b*v_b=0[/tex] <- the negative is because they are in opposite directions

    [tex]m_a*v_a = m_b*v_b[/tex]

    [tex](2m_b)*v_a = m_b*v_b[/tex]


    Now for conservation of energy. The systems total inital energy is 60J. After the exchange is complete all of that energy is kinetic.

    [tex]\frac{1}{2}m_a*(v_a)^2 + \frac{1}{2}m_b*(v_b)^2 = 60[/tex]

    [tex]\frac{1}{2}(2m_b)*(\frac{1}{2}(v_b))^2 + \frac{1}{2}m_b*(v_b)^2 = 60[/tex]

    Now lets simplify that a bit.

    [tex]\frac{3}{2}KE_b=60[/tex] where [tex]KE_b=\frac{1}{2}m_b*(v_b)^2[/tex]
    Last edited: Nov 17, 2005
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