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Momentum problem

  1. Apr 13, 2007 #1
    1. The problem statement, all variables and given/known data

    A car with a gun and a man is rest on a frictionless floor.The total mass is 50m where m is the mass of a single shell.Now,the man fires each shell with a muzzle speed 200 m/s.and the car recoils.What is the speed aftyer the 2nd time firing?

    2. Relevant equations

    v_sc=v_sf-v_cf where v_sc,v_sf,
    v_cf are velocity of shell w.r.t. car,velocity of shell w.r.t. floor, velocity of car w.r.t. floor

    3. The attempt at a solution

    R_CM=[1/(50m)][49m*r_cf+m*r_sf]

    (d/dt)R_CM=[1/(50m)] [49m*v_cf+m* v_sf ]=0

    49m*v_cf+m*[ v_sc + v_cf ] =0 from relevant eqn.

    This gives,v_cf=-[v_sc/50 m/s
    =-200/50 (i) m/s where (i) is the unit vector

    v_sf=(49/50)v_sc

    w.r.t the same frame,
    {48m*v'_cf+m*[v'_sc+v'_cf]}+m*(49/50)v_sc=0
    49v'_cf=-v'_sc-(49/50)v_sc
    v'_cf=-200(1/50+1/49)(i)


    Please check if I went wrong anywhere
     
  2. jcsd
  3. Apr 13, 2007 #2
    I think I am correct.However,I have noticed that even if you do not take the whole of the system in the second case,even then the problem can be done.Just we have to use a more familiar version of conservation of linear momentum and apply it over the region of interest.

    I want to know to what extent it is justified to use two different systems in a SINGLE problem to have a unique result?MY intuition suggests when the results match there should be some deeper way of understanding the physics.
    If the result does not match,...then...it is an accident?
     
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