Two blocks of masses ${\mathrm{m}}_1 \mathrm{and} {\mathrm{m}}_2$ are connected as shown in the figure. The acceleration of the block ${\mathrm{m}}_2$, is:

 

Let a = ${\mathrm{acc}}^{\mathrm{n}} \mathrm{of} {\mathrm{m}}_1; {\mathrm{acc}}^{\mathrm{n}} \mathrm{of} \mathrm{pulley} = \frac{\mathrm{a}+0}{2} = \frac{\mathrm{a}}{2}$

If acceleration of ${\mathrm{m}}_2$ = b

Then 0+$\frac{\mathrm{b}}{2} = \frac{\mathrm{a}}{2}$

Hence a = b

    $\mathrm{T} = {\mathrm{m}}_1\mathrm{a}, {\mathrm{m}}_2\mathrm{g} -\mathrm{T} = {\mathrm{m}}_2\mathrm{a}$

$\therefore \mathrm{a} = \frac{{\mathrm{m}}_2\mathrm{g}}{{\mathrm{m}}_1+{\mathrm{m}}_2}$