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

`m_(1)` = mass of trolley = 200 kg
`m_(2)` = mass of child = 20 kg
`u_(1)` = Intial velocity of trolley = `10 ms^(-1)`
`u_(1)` = Final velocity of trolley
Initial momentum of the system before the child strats running
`= (m_(1)+m_(2))u_(1)`
`= (200 + 20) xx10`
` = 2200 " kg "ms^(-1)`
Now the child starts running with a speed of 4 m `s^(-1)` in a direction opposite to motion of trolley .
` :. ` Child .s velocity relative to trolley ` = 4ms^(-1)` Child .s velocit y relative to ground ` = u_(1) - 4`
Final momentum of the system `=m_(1)u_(1) +m_(2) (u_(1)-4)`
` = 200 u_(1) +20 (u_(1)-4)`
As final momentum = Initial momentum
` 200 u_(1) +20 (u_(1)-4) = 2200`
On solving we get `u_(1) = 10.36 " m "s^(-1)`
time taken by child to cover a disatnce of 10 m = t
= `10/4 = 2.5 s `
Distance covered by trolley in this time = velocity of trolley `xx ` t
` = 10.36 xx2.45`
` = 25.9 m `