Solve the following :

Solution:

Use C.O.L.M,

$m \cdot V_{1}=2 m(0)=m \cdot V_{2}+(2 m)\left(V_{3}\right)$

When B reaches man,

Use C.O.E.L,

$V_{3}=\sqrt{2 g h}$

$\Rightarrow V_{1}-V_{2}=2 V_{3}=2 \sqrt{2 g h}-(1)$

$\frac{1}{2} m V_{1}^{2}=\frac{1}{2} m V_{2}^{2}+\frac{1}{2}(2 m) V_{3}^{2}$

$V_{1}^{2}-V_{2}^{2}=2 V_{3}^{2}=2(2 g h)-(3)$

Solve (1) and (2) to get

$\frac{V_{1}-V_{2}}{V_{1}^{2}-V_{2}^{2}}=\frac{2 \sqrt{2 g h}}{2 \sqrt{2 g h}}=\sqrt{2 g h}$

$V_{1}-V_{2}=2 \sqrt{2 g h}$

$2 V_{1}=3 \sqrt{2 g h}$

$V_{1}=\frac{3}{2} \sqrt{2 g h}$

Use $V_{1}^{2}=u^{2}+2 g h$

$u=\sqrt{V_{1}^{2}-2 g h}$

$=2.5 \sqrt{2 g h}$

Block's initial velocity $>2.5 \sqrt{2 g h}$

$\frac{1}{2} 2 m(0)^{2}-\frac{1}{2} \times 2 m\left(V_{3}^{2}\right)=m g h$