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Homework Help: Change in total kinetic energy in a conservation of momentum problem?

  1. Oct 7, 2005 #1

    On a frictionless, horizontal air table, puck A (with mass [tex]m_1[/tex]) is moving toward puck B (with mass [tex]m_2[/tex]), that is initially at rest. After the collision, puck A has a velocity of [tex]v_1[/tex] to the left, and puck B has velocity [tex]v_2[/tex] to the right.

    1. What was the speed of puck A before the collision?

    My answer to this part of the question was correct. It is [tex](m_2*v_2)/(m_1) - (v_1)[/tex].

    2. Calculate the change in the total kinetic energy of the system that occurs during the collision.

    I think this depends on the first part of the question.

    So, I'm thinking:

    [tex]\Delta K= K_f - K_i[/tex]
    [tex](1/2)(m)(v_f^2 - v_i^2)[/tex]

    I should sum up the velocities, right?

    [tex](1/2)(m_1+m_2)((v_2 - v_1)^2 - ((m_2*v_2)/(m_1) - (v_1))^2)[/tex]

    But, that is not correct.
  2. jcsd
  3. Oct 7, 2005 #2


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    I'm not very clear with what was meant by your last equation. In anycase, you basically do:

    Change in kinetic energy = Kinetic energy of puckA before collision - Kinetic energy of puckA after collision - kinetic energy of puckB after collision.

    KE is a function of only mass and velocity. Since all are known, carefully plugging in the numbers should do the trick.
  4. Oct 7, 2005 #3

    Physics Monkey

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    When calculating kinetic energy, you first square the velocity of each particle, multiply by the particles mass divided by two and then add. You have added the final velocities first and then squared and this is wrong.
  5. Oct 7, 2005 #4
    I inputed [tex](1/2)(m_1)((m_2*v_2)/(m_1) - (v_1))^2 - (1/2)(m_1)(v_1)^2 - (1/2)(m_2)(v_2)^2[/tex] and it was incorrect.

    My understanding of change in total kinetic energy, [tex]\Delta K[/tex], is the difference between the final and initial kinetic energies.

    Physics Monkey, you told me to add. Why is this so?
  6. Oct 7, 2005 #5

    Doc Al

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    [itex]\Delta K = K_f - K_i[/itex], not [itex]K_i - K_f[/itex].
  7. Oct 7, 2005 #6


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    Try applying conservation of momentum and kinetic energy.


    Consider the momentum of the masses to left and right. Remember momentum is proportional to velocity, so it is a vector quantity. Energy is a scalar.
  8. Oct 7, 2005 #7


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    From the looks of it, the problem now isn't with the physics, but rather the math. Try cleaning it up, do things step by step and plug in the numbers. Like what Physics Monkey said, you are doing some incorrect operations by trying to rush the calculation. One by one,

    [tex]KE_1 = \frac{1}{2}m_1v_1^2[/tex]
  9. Oct 10, 2005 #8
    It looks like I definitely did [tex]K_i - K_f[/tex], which obviously is not correct.

    So, I decided to analyze [tex]K_f[/tex] first.

    I got [tex]K_f = (1/2)(m_1)(-v_1)^2 + (1/2)(m_2)(v_2)^2[/tex].

    Then, I decided to analzye [tex]K_i[/tex].

    I got [tex]K_i = (1/2)(m_1)((m_2*v_2)/(m_1) - v_1)[/tex]
    This simplifies to [tex](1/2)(m_2*v_2 - m_1*v_1)[/tex].

    I took the difference [tex]K_f - K_i = (1/2)(m_1)(-v_1)^2 + (1/2)(m_2)(v_2)^2 - (1/2)(m_2*v_2 - m_1*v_1)[/tex] and it was not correct.

    I really think I did everything right this time. ?
  10. Oct 10, 2005 #9

    Doc Al

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    Other than that (which gives the wrong sign) I don't see anything wrong with what you posted in #4.


    You forgot to square the speed.
    Last edited: Oct 10, 2005
  11. Oct 10, 2005 #10
    Thanks so much, Doc Al. I wanted to simplify my answer, but I forgot that it's [tex]v^2[/tex].
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