The blocks A and B shown in the figure have masses ${\text{M}}_{\text{A}}=5\text{kg}\text{and}{\text{M}}_{\text{B}}=4\text{kg}$ . The system is released from rest. The speed of B after A has travelled a distance 1 m along the incline is

 

from the given fig

$4\text{g}-2\text{T}=4\text{a}\to \left(1\right)$

$\text{T}-3\text{g}=5\times 2\text{a}$

$\because \text{T}-5\text{mg}\sin 37°=5\times 2\text{a}\left[\because \sin 37°=\frac{3}{5}\right]$

$\Rightarrow \text{2T}-\text{6g}=\text{20a}\to \left(2\right)$

$\left(1\right)+\left(2\right)$

$-2\text{g}=24\text{a}$

$\text{a}=-\frac{\text{g}}{\text{12}}{\text{ms}}^{-2}$

We know ${\text{v}}^{\text{2}}-{\text{u}}^{\text{2}}=2\text{as}$

${\text{v}}^2=2{\text{a}}_{\text{B}}{\text{s}}_{\text{B}}$

$=2\times \frac{\text{g}}{\text{12}}\times \frac{1}{2}$

${\text{v}}^{\text{2}}=\frac{\text{g}}{\text{12}}$

${\text{v}}_{\text{B}}=\sqrt{\frac{\text{g}}{\text{12}}}=\frac{\sqrt{\text{g}}}{\text{2}\sqrt{\text{3}}}$