Two particles,$1$ and $2$,each of mass $m$,are connected by a massless spring and are on a horizontal frictionless plane,as shown in the figure. Initially,the two particles,with their center of mass at $x_0$,are oscillating with amplitude $a$ and angular frequency $\omega$. Thus,their positions at time $t$ are given by $x_1(t) = (x_0 + d) + a \sin \omega t$ and $x_2(t) = (x_0 - d) - a \sin \omega t$,respectively,where $d > 2a$. Particle $3$ of mass $m$ moves towards this system with speed $u_0 = a \omega / 2$ and undergoes an instantaneous elastic collision with particle $2$ at time $t_0$. Finally,particles $1$ and $2$ acquire a center of mass speed $v_{cm}$ and oscillate with amplitude $b$ and the same angular frequency.
$(1)$ If the collision occurs at time $t_0 = 0$,the value of $v_{cm} / (a \omega)$ will be
$(2)$ If the collision occurs at time $t_0 = \pi / (2 \omega)$,then the value of $4b^2 / a^2$ will be

• A
  $0.75, 4.30$
• B
  $0.75, 4.25$
• C
  $0.75, 4.35$
• D
  $0.75, 4.40$