Here, mass of trolley, $m_1 = 200$ kgspeed of the trolley $v$ = 36 km/h = 10 m/smass of the child, $m_2 = 20$ kgBefore the child starts running, momentum of the system$p_1 = (m_1 + m_2) v$ $= (200 + 20) 10 = 2200\ \mathrm{kg}\,\mathrm{m}\,\mathrm{s}^{-1}$When the child starts running, with a velocity of 4 m/s in a direction opposite to trolley, suppose $v'$ is final speed of the trolley (w.r.t. earth).Obviously, speed of the child relative to earth $(v' - 4)$∴ Momentum of the system when the child is running,$p_2 = 200v' + 20 (v' - 4)$ $= 220v' - 80$As no external force is applied on the system$\therefore p_2 = p_1$$220v' - 80 = 2200$$220v' = 2200 + 80 = 2280$ $v' = \frac{2280}{220} = 10.36\ \mathrm{ms}^{-1}$Time taken by the child to run a distance of 10 m over the trolley,$t = \frac{10\ \mathrm{m}}{4\ \mathrm{ms}^{-1}} = 2.5\ \mathrm{s}$Distance moved by the trolley in this time = Velocity of trolley × time$= 10.36 \times 2.5 = 25.9\ \mathrm{m}$