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(B) Let the particle of mass $m$ be $A$ and the particle of mass $2m$ be $B$. Initially,$A$ moves with velocity $v$ and $B$ is at rest.Applying conser

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$\frac{2\pi R}{v}$

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(B) Let the particle of mass $m$ be $A$ and the particle of mass $2m$ be $B$. Initially,$A$ moves with velocity $v$ and $B$ is at rest.Applying conservation of linear momentum:$mv + 0 = mv_1 + 2mv_2$$v = v_1 + 2v_2$  --- $(i)$Since the collision is elastic,the coefficient of restitution $e = 1$:$e = \frac{v_2 - v_1}{v - 0} = 1$$v_2 - v_1 = v$  --- $(ii)$Adding $(i)$ and $(ii)$:$3v_2 = 2v \implies v_2 = \frac{2}{3}v$Substituting $v_2$ in $(ii)$:$v_1 = v_2 - v = \frac{2}{3}v - v = -\frac{1}{3}v$The negative sign indicates that particle $A$ reverses its direction.The relative velocity of separation is $v_{rel} = v_2 - v_1 = \frac{2}{3}v - (-\frac{1}{3}v) = v$.The distance to be covered for the next collision in the circular tube is the circumference $2\pi R$.Time $t = \frac{	ext{Distance}}{	ext{Relative velocity}} = \frac{2\pi R}{v}$.

$A$ particle of mass $m$ moving with a speed $v$ hits elastically another stationary particle of mass $2m$ in a fixed smooth horizontal circular tube of radius $R$. Find the time when the next collision will take place?

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