Two masses M_1=5 ~kg and M_2=10 ~kg are connected at the ends of an...

Since $M_2 \gt M_1$, therefore $M_2$ moves downwards and $M_1$ moves upwards with an acceleration $a$ as shown in the figure.
$\text{ Free body diagram of }$M_1 $<img src="/api/proxy-image?url=https$ \\ The equation of motion for $M_1$ is T-M_1 g=M_1 a Free body diagram of $M_2<img src="/api/proxy-image?url=https$ \\ The equation of motion for $M_2$ is M_2 g-T=M_2 a Adding (i) and (ii), we get a=$\frac{\left(M_2-M_1\right) g}{M_2+M_1}=\frac{(10-5) g}{(10+5)}=\frac{g}{3}$