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(D) Let the masses be $m$. Since the collision is elastic and the masses are equal,the particles exchange their velocities after the collision.Initial

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$1$

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(D) Let the masses be $m$. Since the collision is elastic and the masses are equal,the particles exchange their velocities after the collision.Initially,the particles have speeds $v$ and $2v$. Let the angular speeds be $\omega$ and $2\omega$. The total angle covered by both particles at the first collision is $2\pi$. Thus,$\omega t_1 + 2\omega t_1 = 2\pi$,which gives $t_1 = \frac{2\pi}{3\omega}$.At this time,the first particle has covered an angle $	heta_1 = \omega t_1 = \frac{2\pi}{3} = 120^\circ$ and the second has covered $240^\circ$.After the first elastic collision,they exchange velocities. Now,the particle that was moving with $2v$ moves with $v$ and vice-versa.For the second collision,the relative angular speed is still $3\omega$,so the time taken is $t_2 = \frac{2\pi}{3\omega}$.In this time,the particle that had velocity $v$ (now $2v$) covers an additional angle of $240^\circ$,and the other covers $120^\circ$.Total angle covered by the particle that started with $v$ is $120^\circ + 240^\circ = 360^\circ$. This particle reaches point $A$ at the second collision.The other particle also reaches point $A$ at the same time. Thus,only $1$ more collision is required after the first one for both to reach point $A$.

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