Solve Momentum Problem: Car w/ Gun & Man Fire Shells at 200 m/s

[neelakash]

Homework Statement

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?

Homework 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

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

 

[neelakash]

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?