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Conservation of momentum

  1. Jan 31, 2005 #1
    The question is here:

    For a...

    for b...

    for c...

    however the problem doesnt say in terms of [tex]x_m[/tex]
    I'm not sure what to do. If I plugged in b, it would give me v=v

    not quite sure
    is that right?
    Last edited: Jan 31, 2005
  2. jcsd
  3. Jan 31, 2005 #2
    the link isn't working, state the problem or fix the link
  4. Jan 31, 2005 #3
  5. Jan 31, 2005 #4
    this is right

    show your work,
    I got a different answer using conservation of energy, I'm assuming m1 and m2 are moving horizontally wrt the ground so gravity is not a factor (its hard to tell from the diagram)

    Evaluating the energy before the collision and at the moment of the springs max compression, i get

    [tex] \frac{1}{2}m_1{v_1}^2 \ + \ \frac{1}{2}m_2{v_2}^2 = \frac{1}{2}k{x_m}^2 \ + \ \frac{1}{2}(m_1 + m_2){v_f}^2 [/tex]

    substituting the value calculated in (a) for [itex]v_f[/itex], this becomes

    [tex] \frac{1}{2}m_1{v_1}^2 \ + \ \frac{1}{2}m_2{v_2}^2 = \frac{1}{2}k{x_m}^2 \ + \ \frac{1}{2}(m_1 + m_2) \frac{(m_1 v_1 + m_2 v_2)^2}{(m_1 + m_2)^2} [/tex]

    after all the algebra and simplifying my result is

    [tex] x_m = \sqrt{ \frac{ m_1 m_2 (v_1 - v_2)^2}{k(m_1 + m_2)} } [/tex]

    I think this is right although I'm not %100 sure since I don't have my book with me

    It looks like you tried to use energy... The thing you have to keep in mind when using conservation of energy, is that you have to evaluate the whole system. In this case m1 and m2 as well as the spring are part of the system, so you can't break them up. Your equation says that all the energy from the spring goes in to the kinetic energy of m1, but this isn't true. The spring pushes on m1, but by newton's 3rd law, m1 pushes equally hard on the spring. m2 is connected to the spring, so its energy is affected too.

    I treated this as another conservation of momentum problem although now starting from the point when both masses are stuck together

    letting [itex]{v_1}_i [/itex] & [itex]{v_1}_f[/itex] be the velocity of m1 before and after the collision respectifully ( same notation goes for m2 ) and letting [itex] v_f [/itex] indicate the velocity calculated in part a, I get

    [tex] (m_1 + m_2)v_f = m_1 {v_1}_f + m_2 {v_2}_f [/tex]

    [tex] (m_1 + m_2)\frac{m_1 {v_1}_i + m_2 {v_2}_i}{m_1 + m_2} = m_1 {v_1}_f + m_2 {v_2}_f [/tex]

    solving for [itex] {v_1}_f [/itex] i get

    [tex] {v_1}_f = {v_1}_i + \frac{m_2({v_2}_i - {v_2}_f)}{m_1} [/tex]

    the same can be done for [itex] {v_2}_f [/itex]

    I think this is the answer, although I am not %100 sure
    Last edited: Feb 1, 2005
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