d
(d) Let the velocity acquired by \(A\) and \(B\) be \(V\), then
\(mv = mV + mV \)

\(\Rightarrow V = \frac{v}{2}\)
Also \(\frac{1}{2}m{v^2} = \frac{1}{2}m{V^2} + \frac{1}{2}m{V^2} + \frac{1}{2}k{x^2}\)
Where \( x\) is the maximum compression of the spring. On solving the above equations, we get \(x = v{\left( {\frac{m}{{2k}}} \right)^{1/2}}\)
At maximum compression, kinetic energy of the
\(A -B\) system \( = \frac{1}{2}m{V^2} + \frac{1}{2}m{V^2} = m{V^2} = \frac{{m{v^2}}}{4}\)