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 trolley, `m_(1)=200kg, ` speed of the trolley , `upsilon=36km//h =10m//s, ` mass of the child, `m_(2)=20kg `
Before the child starts running, momentum of the system
`p_(1)=(m_(1)+m_(2))upsilon=(200+20)10=2200kgms^(-1)`
When the child starts running, with a velocity of `4m//s` in a direction opposite to trolley, suppose `upsilon^(')` is final speed of the troley (w.r.t. earth). Obviously, speed of the child relative to earth `=(upsilon^(')-4)`
`:. ` Momentum of the system when the child is running,
`p_(2)=200upsilon^(')+20(upsilon^(')-4)=220upsilon^(')-80`
As no external force is applied on the system `:. p_(2)=p_(1)`
`200upsilon^(')-80=2200`
`220upsilon^(')=2200+80=2280`
`upsilon^(')=(2280)/(220)=10.36ms^(-1)`
Time taken by the child to run a distance of 10m over the trolley, `t=(10m)/(4ms^(-1))=2.5s`
Distance moved by the trolley in this time `=` velocity of trolley `xx` time `=10.36xx2.5=25.9m`