Two masses M _ 1 and M _ 2 are tied together at the two ends of a...

In both cases ${m}_{1} > {m}_{2}$. Therefore, ${m}_{1}$ will accelerate downward and ${m}_{2}$ will accelerate upward.

For ${m}_{1}$,

${m}_{1}g - T = {m}_{1}a$

For ${m}_{2}$,

$T - {m}_{2}g = {m}_{2}a$

Adding the above equations we get,

$a = \dfrac{{m}_{1} - {m}_{2}}{{m}_{1} + {m}_{2}}g$

$⇒ {a}_{1} = \dfrac{{m}_{1} - {m}_{2}}{{m}_{1} + {m}_{2}}g = \dfrac{2{m}_{2} - {m}_{2}}{2{m}_{2} + {m}_{2}}g = \dfrac{g}{3}$

and

$⇒ {a}_{2} = \dfrac{{m}_{1} - {m}_{2}}{{m}_{1} + {m}_{2}}g = \dfrac{3{m}_{2} - {m}_{2}}{3{m}_{2} + {m}_{2}}g = \dfrac{g}{2}$

$⇒ \dfrac{{a}_{1}}{{a}_{2}} = \dfrac{2}{3}$