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Explosions,Center of Mass

  1. Jul 3, 2015 #1
    1. The problem statement, all variables and given/known data
    2jectae.jpg

    [itex] 2 [/itex] masses connected with a solid rod .distace between them is [itex] 2L [/itex] ,
    total mass is [itex]M[/itex] .

    [itex] B [/itex] total mass is [itex]m[/itex] ,length [itex] L [/itex] can freely move between [itex]A[/itex] and [itex]A'[/itex]
    on flat and frictionless surface. all collisions are elastic.

    at the beginning [itex] B[/itex] was stuck to [itex] A [/itex] and all the system was moving to the right ,[itex] v_0[/itex] and kinetic energy [itex] T_0[/itex] .
    at some moment ,explosion take place and the system get additional kinetic energy [itex]T[/itex] .

    1.what is the velocity of [itex]B[/itex] ,[itex] v_B[/itex] as function of [itex] v_0,T,m,M[/itex] ?
    2.what is the velocity of [itex]A+A' [/itex] ,immediately after the explosion ?
    3.what is the relation between [itex]T_0[/itex] and [itex]T[/itex] ,if [itex]A [/itex] don't move to the left ?
    4.after the explosion [itex]B [/itex] start to move between [itex] A[/itex] and [itex]A' [/itex],what is the time between the
    collision of [itex]B [/itex] with [itex]A [/itex] , and the collision between [itex]B [/itex] and [itex]A' [/itex] ?


    2. Relevant equations
    The total kinetic energy (in the lab frame) is :
    [itex] K = K_{cm} +\frac{1}{2} \sum_{n=0}^N m_i (v'_i)^2 [/itex]


    3. The attempt at a solution
    before the explosion:
    [itex] T = \frac{1}{2}(M+m)v_{cm}^2 + 0 +0 [/itex] ,given [itex] v_0 = v_{cm}\hat x [/itex] we get [itex] T = \frac{1}{2}(M+m)v_0^2 + 0 +0 [/itex]

    after :
    [itex] T+ T_0 = \frac{1}{2}(M+m)v'^2_{cm} + \frac{1}{2}Mv'^2_A +\frac{1}{2}mv'^{2}_B [/itex]
    or
    [itex]
    \frac{1}{2}(M+m)v_0^2+ T_0 = \frac{1}{2}(M+m)v'^2_{cm} + \frac{1}{2}Mv'^2_A +\frac{1}{2}mv'^{2}_B [/itex]

    [itex] v'_{cm}[/itex] = is the velocity of center of mass after the explosion ..so it move the same velocity
    as before ,which mean - [itex] v'_{cm} = v_0[/itex] .

    so we get :
    (1) [itex] 0 = Mv'_A +mv'_B [/itex] - momentum
    (2) [itex] T_0 = \frac{1}{2}Mv'^2_A +\frac{1}{2}mv'^{2}_B [/itex]
    i get that if
    [itex] m - \frac{m^2}{M} > 0 [/itex] then [itex] v'_B = \pm\sqrt{\frac{2T_0}{m - \frac{m^2}{M} }}\hat x'[/itex]
    and [itex] v_B = v_0 \hat x+ ( \pm\sqrt{\frac{2T_0}{m - \frac{m^2}{M} }}) \hat x[/itex]


    Q:
    1. is this the solution ?
    2.if 1 is o.k then [itex]v_b [/itex] ...easy
    3 - what it means "move to the left " ? ...at c.m reference system ,it`s look to me that 2 masses cant move
    to the same direction ,do they ? if so , (1) - wrong solution .
    any way don't have a clue ..
    4.??
     
  2. jcsd
  3. Jul 3, 2015 #2

    SammyS

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    Staff Emeritus
    Science Advisor
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    What you have given as T above is actually T0.
    What you have as T0 on the left side of the above equation is actually T. The ##\displaystyle \ \frac{1}{2}(M+m){v_0}^2 \ ## is T0.
    See what making those correction will do.
     
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