Two point masses `m_1` and `m_2` are connected by a spring of natural length `l_0`. The spring is compressed such that the two point masses touch each other and then they are fastened by a string. Then the system is moved with a velocity `v_0` along positive x-axis. When the system reached the origin, the string breaks `(t=0)`. The position of the point mass `m_1` is given by `x_1=v_0t-A(1-cos omegat)` where A and `omega` are constants. Find the position of the second block as a function of time. Also, find the relation between A and `l_0`.

Explanation

(a) As no external force is acting on the system, hence linear momentum of the system is conserved
i.e., `m_(1)(dx_(1))/(dt)+m_(2)(dx_(2))/(dt)=(m_(1)+m_(2))v_(0)`
or `m_(2)(dx_(2))/(dt)=(m_(1)+m_(2))v_(0)-m_(1)(dx_(1))/(dt)`
`=(m_(1)+m_(2))v_(0)=m_(1)(d)/(dt)[v_(0)t-A(1-cosomegat)]`
`=(m_(1)+m_(2))v_(0)-m_(1)(v_(0)t-Aomegasingomegat)`
`=m_(2)v_(0)+m_(1)Aomegasinomegat`
Position of particle of mass `m_(2)` at time t is
`x_(2)=int_(0)^(t)v_(0)dt+int_(0)^(t)(m_(1)Aomega)/(m_(2))sinomegadt`
`=v_(0)+(m_(1)A)/(m_(2))(1-cosomegat)`
(b) `x_(2)-x_(1)=[v_(0)t+(m_(1)A)/(m_(2))(1-cosomegat)]`
`-[v_(0)t-A(1-cosomegat)]`
`=(m_(1)A)/(m_(2))(1-cosomegat)+A(1-cosomegat)`
Since the string breaks, the separation `(x_(2)-x_(1))`
between the two blocks will be equal to `l_(0)` i.e., maximum value of `x_(2)-x_(2)=l_(0)`. It will be so if `cosomegat=0`
`:. l_(0)=(m_(1)A)/m_(2)+A=((m_(1)+m_(2))/(m_(2)))`