Block A of mass 2kg moving with  10 m/s strikes a spring of constant  ${\text{π}}^2\text{\hspace{0.17em}N}/\text{m}$ attached to another 2kg block  B at rest kept on a smooth floor. The time for which rear moving block  A remain in contact with spring will be

detailed solution

Correct option is C

According to the statementm1=m2=2 kg We have a relation for the time period of two spring mass systemt=2πμk.....1 Where  k=spring constant= π2 N/mμ= reduced mass= m1m2m1+m2The time period of both oscillations are same2πm1k1=πm1k2....2 Now calculations for  μ=m1m2m1+m2Substituting the valuesμ=2×22+2=44=1 For  μ=1Substituting the value of 'μ' & 'k' in (1)T=2π1π2  T=2 sBut it takes  T2 time for first block to detach, implies time T2=0.5×2=1 sThe time for which rear moving block remain in contact with spring will be 1 second.