M is a fixed wedge. Masses ${\mathrm{m}}_1 \mathrm{and} {\mathrm{m}}_2$ are connected by a light string. The wedge is smooth and the pulley is smooth and fixed. ${\mathrm{m}}_1 =$l0 kg and $m_2 =$ 7.5 kg. When $m_2$ is just released, the distance it will travel in 2 seconds is

 

F.B.D of ${\mathrm{m}}_1 \mathrm{and} {\mathrm{m}}_2$

Equation of motion

For  ${\mathrm{m}}_1 : \mathrm{T} -{\mathrm{m}}_1\mathrm{g}\left(\frac{1}{2}\right) = {\mathrm{m}}_1\mathrm{a}-----\left(\mathrm{i}\right)$

For ${\mathrm{m}}_2 : {\mathrm{m}}_2\mathrm{g} -\mathrm{T} = {\mathrm{m}}_2\mathrm{a}------\left(\mathrm{ii}\right)$

From (i) and (ii), a = $\frac{({\mathrm{m}}_2-{\displaystyle \frac{{\mathrm{m}}_1}{2}})}{({\mathrm{m}}_1+{\mathrm{m}}_2)}\mathrm{g} = \frac{10}{7}\mathrm{m}/{\mathrm{s}}^2$

Hence distance travelled by block ${\text{m}}_2$ in 2 sec; using

$\mathrm{s} = \mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

$\mathrm{s} = 0\times 2+\frac{1}{2}\left(\frac{10}{7}\right)\times {\left(2\right)}^2 = \frac{20}{7}\mathrm{m} = 2.8 \mathrm{m}$