A trolley of mass 200 kg moves with a uniform speed of 36 km / h on frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of `mu m s^(-1)` relative to the trolley in a direction opposite to the trolley in a direction opposite to the its motion and jumps out of the trolley. What is the final speed of the trolley ? How much has been trolley moved from the time the child begins to run ?

Explanation

Here, mass of trolley, `m_(1) = 200kg`
speed of the trolley v = 36 km/h = 10m/s
mass of the child , `m_(2) = 20 kg`
Before the child starts running , momentum of the system
`P_(1) = (m_(1) + m_(2)) v = (200 + 20) 10 = 220 kg ms^(-1)`
`:.` Momentum of the system when the child is running
`P_(2) = 200v^(1) + 20 (v^(1) - 4) = 220v^(1) - 80`
As no external force is applied on the system
`:. P_(2) = P_(1)`
` 220v^(1) - 80 = 2200`
`220v^(1) = 2200 + 80 = 2280`
`v^(1) = (2280)/(220) = 10.36 ms^(-1)`
Time taken by the child to run a distance of
`10 m `over the trolley , t `= (10 m)/( 4 ms^(-1) = 2.5 s`
Distance moved by the trolley in this time = Velocity of trolley `xx` time ` = 10.36 xx 2.5 = 25.9 `m