# Homework Help: Momentum problem revisited

1. Apr 23, 2007

### neelakash

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

Conservation of Linear momentum

3. The attempt at a solution

I did the problem in these two ways---Both are correct.Now,in the first problem, I have used car+1st shot bullet+2nd shot bullet as the system.

Whereas in the 2nd method,I used two different systems per shot--- composed of the car+ one bullet.Then,I added the answers vectorially
.Please let me know why this method gives correct result.

1st method:
R_CM=[1/(50m)][49m*r_cf+m*r_sf]
(d/dt)R_CM=[1/(50m)] [49m*v_cf+m* v_sf ]=0

[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]

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
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)

second Method:

Let an object have mass M and eject a smaller mass m with a relative velocity v in the negative direction. What is the change in the velocity (dv) of M? P_i = P_f ---> Mv_i = (M - m)v_f + m(v_f - v) = Mv_f - mv ---> M(v_f - v_i) = mv ---> Mdv = mv ---> dv = (m/M)v.

Using this general relation, dv_1 = (1/50)*200, and dv_2 = (1/49)*200. Since the initial velocity was 0, the final velocity of M is 0 + dv_1 + dv_2 = (1/50 + 1/49)*200.

2. Apr 23, 2007

### neelakash

I hope I understand now.Since as an observer I always see all of the elements of the system---car,1st and 2nd bullets,I will see the CM at rest.So,the 1st method is appropriate.I was unable to visualise a different system when I am seeing all at a time.OK,I can select another system...as physics remains unchanged.However,in this system,the CM will be moving.

So,selecting (car+2nd bullet) as the system,we may avoid the 1st bullet term,but the CM motion term cannot be avoided.Whereas if I use a single system all the way,I will have to include the 1st bullet term,but there will be no CM motion term.