A flat car of mass `m` is at rest on a frictionless floor with a child of mass `m` standing at its edge. If the child jumps off from the car towards right with an initial velocity `u`, with respect to the car, find the velocity of the car after its jump.
A flat car of mass `m` is at rest on a frictionless floor with a child of mass `m` standing at its edge. If the child jumps off from the car towards right with an initial velocity `u`, with respect to the car, find the velocity of the car after its jump.
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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 the 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 the 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))`
Velcoity 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)+Mu)/(M+m)`
According to momentum conservation.
`(M+m)v_(1)=Mv_(2)+m(u+v_(2))`
Velcoity 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)+Mu)/(M+m)`
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