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⇒a=‌

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 x1 and x2, respectively. So, on taking the frame of centre of mass there will act a pseudo force equal to ma on the masses.
On applying the work-energy theorem, taking the centre of mass as a reference point.
W‌all forces ‌=∆K
[∵ w.r.t. centre of mass ∆K=0]
⇒‌‌x1+x2=‌

F

K

⇒x‌total ‌=‌

F

K

Here, F=10N,K=2500N/m2
So, So, the maximum distance between the blocks will be = natural length +x‌total ‌
=10cm+0.4cm=10.4cm
[∵ Natural length =10cm (given)]