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JEE Advance 2024
Paper-2 2024
Question
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 $x_2(t)=\left(x_0-d\right)-\mathrm{a} \sin \omega t$, respectively, where $d>2 a$. 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 $\omega$. $$ \text { If the collision occurs at time } t_0=0 \text {, the value of } v_{\mathrm{cm}} /(a \omega) \text { will be } $$
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