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 m s^–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?

Given,

Mass of the trolley, m = 200 kg

Speed of the trolley, v=36km/h=(36×1000 m)/(60×60 s)= 10m/s

Mass of the child, m’ = 20 kg

Initial momentum of the system involving the child and the trolley, p_i = (m+m’)v

⇒ p_i = (200 kg + 20 kg)×10 ms^-1

⇒ p_i = 2200 kg m s^-1

Let v’ be the final velocity of the trolley with respect to the ground.

Final velocity of the child with respect to ground, v’’ = Final velocity of trolley with respect to ground – final velocity of child with respect to trolley.

So, v’’ = v’ – 4

Final momentum of the system, p_f = mv’ + m’v’’

⇒ p_f = mv’ + m’(v’-4)

⇒ p_f = 200v’ + 20(v’-4)

⇒ p_f = 220v’-80

According to the law of conservation of linear momentum,

Initial momentum (p_i) of system = Final momentum (p_f) of system

⇒ 2200 = 220v’-80

⇒ 220v’ = 2280

⇒ v’ = 2280/220

⇒ v’ = 10.36 m/s

Length of the trolley, l = 10 m

Speed of the child with respect to trolley, v’ = 4m/s

Time taken by the boy to cover the distance, t = l/v’

⇒ t = 10m/4 ms^-1

⇒ t = 2.5 s

Distance moved by the trolley in this time, s = v’t

⇒ s = 10.36 ms^-1 × 2.5 s

⇒ s = 25.9 m