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Motion of two constrained masses

  1. Feb 20, 2015 #1
    1. The problem statement, all variables and given/known data
    The attached images shows everything. ([itex]{\bf{e}}_1[/itex] denotes the direction of X and [itex]{\bf{e}}_2[/itex] denotes the direction of Y).

    Initially, the spring is force free when [itex]X_{0}=0.4 m[/itex] (which yields [itex]Y_{0}=0.3 m[/itex]). Also, at this instant, the velocity of B is [itex]{\bf{V}}_{B0}=-v_{0}{\bf{e}}_{2}[/itex] such that the corresponding velocity of A is [itex]{\bf{V}}_{A0}=v_{0}(Y/X){\bf{e}}_{1}=0.75v_0{\bf{e}}_{1}[/itex].

    I'm asked to find the minimum value of [itex]v_0[/itex] for which B hits the floor.

    2. Relevant equations
    Energy conservation yields
    [tex](\frac{1}{2}m_{B}V_B^{2}+\frac{1}{2}m_{A}V_A^{2})-(\frac{1}{2}m_{B}V_{B0}^{2}+\frac{1}{2}m_{A}V_{A0}^{2})+\frac{1}{2}k(X_{0}-L)^2=0[/tex]
    However, the kinematic constrains (L is constant) yields
    [tex]XV_{A}+YV_{B}=0[/tex]
    So, when B hits the floor, [itex]Y=0[/itex] so that, at this instant, [itex]V_{A}=0[/itex]. Hence,
    [tex]V_B^{2}=(V_{B0}^{2}+\frac{m_{A}}{m_{B}}V_{A0}^{2})-\frac{k}{m_B}(X_{0}-L)^2=v_{0}^{2}(1+\frac{2}{3}(0.75)^{2})-\frac{100}{3}(0.4-0.5)^2[/tex]
    Now, my requirement is that [itex]V_B=0[/itex] when it hits the floor, so that the last equation yields
    [tex]v_{0}\approx0.49 m/s[/tex]
    while the right answer is
    [tex]v_{0}\approx0.37 m/s[/tex]
    Am I doing/assuming something wrong here?

    I would appreciate your help!

    3. The attempt at a solution
    See the previous item.
     

    Attached Files:

  2. jcsd
  3. Feb 20, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    While it would help if you show more steps, I agree with your result. 0.37 m/s would be the speed of A.
     
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