Two persons each of mass m are standing at the two extremes of a railroad car of mass M resting on a smooth track figure. The person on left jumps to the left with a horizontal speed u with respect to the state of the car before the jump. Thereafter, the other person jumps to the right, again with the same horizontal speed u with respect to the state of the car before his jump. Find the velocity of the car after both the person have jumped off.

Explanation

Let car attain the velocity v in right ward and velocity of man A with respect to ground is v' then
v'=v-u
from momentum conservation
0 = mv' + (M + m)v
`rArr` m(v – u) + (M + m)v = 0
`rArrv=(mu)/((M+2m))`
After wards mass B jumps to the right with the same horizontal speed u with respect to car, than car attain v" velocity from linear momentum conservation.
`(M+m)v=m(u+v")+Mv"`
`(M+m)[(mu)/(M+2m)]=mu+(m+M)v"`
Now `v"=(m^(2)u)/((M+2m)(M+m))`