A flat car of mass `M` with a child of mass `m` is moving with a velocity `v_(1)`. The child jumps in the direction of motion of car with a velocity `u` with respect to the car. Find the final velocities of the child and that of the car after jump.

Explanation

This case is similar to the previous example, except now the car is moving before jump. Here also no external force is acting on the system in horizontal direction, hence momentum remains conserved in this direction. After jump car attains a velocity `v_(2)` in the same direction, which is less than `v_(1)`, due to backward push of the child for jumping. After jump child attains a velocity u + `v_(2)` in the direction of motion of car, with respect to ground.

According to momentum conservation
`(M+m)v_(1)=Mv_(2)+m(u+v_(2))`
Velocity of car after jump is
`v_(2)=((M+m)v_(1)-mu)/(M+m)`
Velocity of child after jump is `u+v_(2)=((M+m)v_(1)+(M)u)/(M+m)`