TS EAMCET 14-Sep-2020 Shift 1 Solved Paper

In the case of maximum elongation, whole system will be moving with a common acceleration $a$, so for the system $F=(m+m)a\Rightarrow a=\;\frac{F}{2m}$...(i) The centre of mass of the system will also move with the same acceleration $a$. Let us suppose that both the masses elongate the spring by a distance $x_{1}$ and $x_{2}$, respectively. So, on taking the frame of centre of mass there will act a pseudo force equal to $m a$ on the masses. On applying the work-energy theorem, taking the centre of mass as a reference point. $W_{\;\text{all forces }\;}=\Delta K$ $[ \because$ w.r.t. centre of mass $\Delta K=0]$ $\Rightarrow \;\;x_{1}+x_{2}=\;\frac{F}{K}\Rightarrow x_{\;\text{total }\;}=\;\frac{F}{K}$ Here, $F=10\ \text{N},\ K=2500\ \text{N}/\text{m}^{2}$ So, So, the maximum distance between the blocks will be $=$ natural length $+x_{\;\text{total }\;}$ $=10\ \text{cm}+0.4\ \text{cm}=10.4\ \text{cm}$ $[ \because$ Natural length $=10\ \text{cm}$ (given)]