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 $d>2 a$. 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 $v_{ cm }$ 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 $4 b^2 / a^2$ will be

Give the answer or quetion ($1$) and ($2$)

224490-q

  • [IIT 2024]
  • A

    $75,4.30$

  • B

    $75,4.25$

  • C

    $75,4.35$

  • D

    $75,4.40$

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