A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a 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 `4 ms^(-1)` relative 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 the trolley moved from the time the child begins to run?

Explanation

Mass of trolley `m_(1)= 200 kg`
Speed of trolley `v = 36 kmh^(-1)`
` = (36 xx1000)/(3600)`
` = 10 ms^(-1)`
Mass of child `m_(2) = 20 kg`
If momentum (Child +trolley) of boy before running is `p_(i)` , then
`p_(i) =(m_(1)+m_(2))v`
`= (200+20) xx10`
`= 2200 kgms^(-1) " "....(1)`
Let v. be the final velocity of thr trolley with respect to the ground .
Final velocity of the boy with respect to the ground = `v. - 4`
Final momentum ,
`p_(f) =m_(1)v+20(v-4)`
`= 200 v+20(v-4)" "...(2)`
Since no external force is exerted the momentum is conserved ,
From equation (1) and (2) ,
`p_(i) =p_(f)`
`2200 = 200 v +20 v - 80`
` :. 2280 = 220 v `
` :. v = (2280)/(220) = 10.36 ms^(-1)`
If time taken by child to run on trolley upto distance 10 m is t. then
`t = (10)/4 = 2.5 s `
If the distance moved by the trolley in this time
x = speed of trolley `xx` time ,
` = v xxt `
` = 10.36 xx2.5`
` = 25.9 `