A particle of mass m moving with a speed v hi | vedclass

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Question

Particle $\mathrm{B}$ is a rest

$\mathrm{mv}+0=\mathrm{mv}_{1}+2 \mathrm{mv}_{2}$

$	herefore \mathrm{v}=\mathrm{v}_{1}+2 \mathrm{v}_{2}$

Vedclass · Accepted Answer

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

Vedclass · Answer

Particle $\mathrm{B}$ is a rest

$\mathrm{mv}+0=\mathrm{mv}_{1}+2 \mathrm{mv}_{2}$

$	herefore \mathrm{v}=\mathrm{v}_{1}+2 \mathrm{v}_{2}$        $...(i)$

$\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{v}+0}=1$

$\Rightarrow \mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{v}$          $...(ii)$

Adding $( i )$+$(ii)$

$3 v_{2}=2 v$

$	herefore \mathrm{v}_{2}=\frac{2}{3} \mathrm{v} \quad \mathrm{v}_{1}=\mathrm{v}_{2}-\mathrm{v}=\frac{-\mathrm{v}}{3}$

$\mathrm{Now},(\mathrm{ii})+(\mathrm{iv})$

$t=\frac{2 \pi r}{v_{2}-v_{1}}=\frac{2 \pi r}{\frac{2 v}{3}+\frac{v}{3}}$

$t=\frac{2 \pi r}{v}$

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

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