How Does Firing Shells Affect the Speed of a Recoiling Car?

[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

Conservation of Linear momentum

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.

 

[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.

 

[m2003]

This is the same result as the first method.

I would commend the student for using two different methods to solve the problem and obtaining the same correct result. It shows a good understanding of the concept of conservation of linear momentum and the ability to apply it in different ways.

In the first method, the student has correctly used the system of car+1st shot bullet+2nd shot bullet and applied the conservation of linear momentum equation to find the final velocity.

In the second method, the student has used a general relation to find the change in velocity of the car after each shot and then added the two changes to find the final velocity. This method is also valid and gives the same result, showing the versatility of the concept of conservation of linear momentum.

Overall, both methods are correct and it is good to see the student exploring different approaches to solve the problem. Keep up the good work!