Paragraph II: 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…

Question

Paragraph II: 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 $v_{\mathrm{cm}}$ 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 ________ .

Aicharya · Accepted Answer

At $\mathrm{t}_0=0$ Before collision After collision $\begin{aligned} \mathrm{v}_{\mathrm{CM}} & =\frac{\mathrm{m} \cdot \frac{\mathrm{a} \omega}{2}+\mathrm{m} \cdot \mathrm{a} \omega}{\mathrm{m}+\mathrm{m}} \ \mathrm{v}_{\mathrm{CM}} & =\frac{3 \mathrm{a} \omega}{4} \ \frac{\mathrm{V}_{\mathrm{CM}}}{\mathrm{a} \omega} & =\frac{3}{4} \ \frac{\mathrm{V}_{\mathrm{CM}}}{\mathrm{a} \omega} & =0.75\end{aligned}$ ;

Paragraph II: Two particles, 1 and 2 , each of mass $m$, are connected by a massless spring, and are on a…

Solution