$A$ 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 $4\; m/s$ relative to the trolley in a direction opposite to 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?

(C) Mass of the trolley,$M = 200\; kg$.
Initial speed of the trolley,$v = 36\; km/h = 10\; m/s$.
Mass of the boy,$m = 20\; kg$.
Initial momentum of the system (boy + trolley) $= (M + m)v = (200 + 20) \times 10 = 2200\; kg\; m/s$.
Let $v'$ be the final velocity of the trolley with respect to the ground.
Final velocity of the boy with respect to the ground $= v' - 4$.
Final momentum $= Mv' + m(v' - 4) = 200v' + 20v' - 80 = 220v' - 80$.
According to the law of conservation of momentum,Initial momentum = Final momentum:
$2200 = 220v' - 80$.
$220v' = 2280$.
$v' = \frac{2280}{220} \approx 10.36\; m/s$.
Length of the trolley,$l = 10\; m$.
Speed of the boy relative to the trolley,$u = 4\; m/s$.
Time taken by the boy to run,$t = \frac{l}{u} = \frac{10}{4} = 2.5\; s$.
Distance moved by the trolley $= v' \times t = 10.36 \times 2.5 = 25.9\; m$.