A trolley having mass of 200 kg moves with uniform speed of 36 kmh...

since, no external force is acting on the system, we can apply conservation of linear momentum.

Speed of $200$ kg trolley $= \dfrac{36 \times 1000}{60 \times 60} = 10ms^{- 1}$

If $u$ be the initial velocity of trolley, ${v}_{b}$ be the absolute velocity of the boy after the beginning of journey of the boy, their relative velocity is $4$.

So, $v^{'} - {v}_{b} = 4 ⇒ {v}_{b} = \left(v^{'} - 4\right)$

Now, applying law of conservation of momentum,

Momentum before the boy begins tu run
= Momentum after the beginning of boy's running

$⇒ 220 \times10 = 200v^{'} + 20 \left(v^{'} - 4\right)$

$2200 = 220v^{'} - 80 ⇒ 220v^{'} = 2280$

$v^{'} = \dfrac{2280}{220} =10. 36ms^{- 1}$