JEE Advanced 2024 · Paper 2, Physics question 17

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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 instantaneous elastic collision with particle 2, at time $t_0$. Finally, particles 1 and 2 acquire a center of mass speed $v_{\text{cm}}$ and oscillate with amplitude $b$ and the same angular frequency $\omega$.