A trolly of mass 200kg moves with a uniform speed of 36 km`//`h on a frictionless track. A child of mass 20kg runs on the trolly from one end to the other (10m away) with a speed of `4m//s` relative to the trolly in a direction opposite to the trolly's motion and jumps out of the trolly. How much has teh trolly moved from the time the child begins to run ?

Explanation

Here, mass of trolly, `m_(1)=200kg`
speed of trolly, `v=36km//h`
`=(36xx1000)/(60xx60)m//s=10m//s`
mass of chid, `m_(2)=20 kg`
Before the child starts running, linear momentum of the system
`P_(1)=(m_(1)+m_(2))v=(200+20)10=2200kg m//s`
When the child starts running with a velocity of `4m//s` in a direction opposite to trolly, suppose `v^(')` is final speed of the troly w.r.t. earth. Speed of child relative to earth
`=v^(')-4`
Momentum of the system when the child is running, `p_(2)=200v^(')+20(v^(')-4)=220v^(')-80`
As no external force is applied ono teh system,
`p_(2)=p_(1)`
`220v^(')-80=2200`
`220v^(')=2200+80=2280`
`v^(')=(2280)/(220)=10.36m//s`
Time taken by the child to run a distance of 10m over the trolly.
`t=(10m)/(4m//s)=2.5s`
Distance moved by the trolly in this time
`=` velocity of trolly `xx` time
`=10.36xx2.5`
`=25.9m`