JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Conservation of Momentum

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 $x_1(t)=\left(x_0+d\right)+a\sin \omega t$ and $x_2(t)=\left(x_0-d\right)-a\sin \omega t$, respectively, where $d>2a$. Particle 3 of mass $m$ moves towards this system with speed $u_0=a\omega /2$, and undergoes instantaneous elastic collision with particle 2, at time $t_0$. Finally, particles 1 and 2 acquire a center of mass speed $v_{\mathrm{cm}}$ and oscillate with amplitude $b$ and the same angular frequency $\omega$.

If the collision occurs at time $t_0=0$, the value of $v_{\mathrm{cm}}/(a\omega )$ will be ____.