Step-by-Step Solution

Let the two particles be P1 and P2, with equal mass m. They move in a horizontal circular orbit of radius R. Let's assume point A is at an angular position θ=0.

Initial State (t=0):

• P1 starts at A with tangential velocity v. Let's assume it moves counter-clockwise (CCW). Its angular velocity is $ω_1 = v/R$.
• P2 starts at A with tangential velocity 2v in the opposite direction (clockwise, CW). Its angular velocity is $ω_2 = 2v/R$.

1. First Collision

• The particles move towards each other. Their relative angular speed is the sum of their individual angular speeds: $ω_{rel} = ω_1 + ω_2 = \frac{v}{R} + \frac{2v}{R} = \frac{3v}{R}$
• The time $t_1$ until their first collision can be found by considering that they must cover a total angle of 2π relative to each other: $t_1 = \frac{2\pi}{ω_{rel}} = \frac{2\pi}{3v/R} = \frac{2\pi R}{3v}$
• To find the location of the first collision, let's calculate the angle $θ_{C1}$ covered by P1 in time $t_1$: $θ_{C1} = ω_1 \times t_1 = \frac{v}{R} \times \frac{2\pi R}{3v} = \frac{2\pi}{3}$
• So, the first collision occurs at an angular position of 2π/3 from point A.

2. State after the First Collision

• The problem states the collisions are elastic. For two particles of equal mass undergoing a one-dimensional elastic collision, they exchange their velocities.
• Before the collision, P1 has speed v (CCW) and P2 has speed 2v (CW).
• After the collision, their velocities are exchanged:
  • P1's new velocity is 2v (CW). Its angular velocity becomes $ω'_1 = 2v/R$ (CW).
  • P2's new velocity is v (CCW). Its angular velocity becomes $ω'_2 = v/R$ (CCW).

3. Second Collision

• After the first collision at $θ_{C1} = 2π/3$, the particles again move towards each other.
• Their new relative angular speed is: $ω'_{rel} = ω'_1 + ω'_2 = \frac{2v}{R} + \frac{v}{R} = \frac{3v}{R}$
• The time interval $Δt_2$ between the first and second collision is: $Δt_2 = \frac{2\pi}{ω'_{rel}} = \frac{2\pi}{3v/R} = \frac{2\pi R}{3v} = t_1$
• The location of the second collision $θ_{C2}$ can be found by calculating the angular displacement of one particle from the first collision point. Let's track P2, which moves CCW with angular speed $ω'_2 = v/R$: $Δθ_2 = ω'_2 \times Δt_2 = \frac{v}{R} \times \frac{2\pi R}{3v} = \frac{2\pi}{3}$
• The position of the second collision is: $θ_{C2} = θ_{C1} + Δθ_2 = \frac{2\pi}{3} + \frac{2\pi}{3} = \frac{4\pi}{3}$

4. State after the Second Collision

• The velocities are exchanged again.
  • P1's new velocity is v (CCW). Its angular velocity is $ω_1 = v/R$.
  • P2's new velocity is 2v (CW). Its angular velocity is $ω_2 = 2v/R$.
• This configuration of velocities is identical to the initial state at t=0.

5. Next Meeting Point

• The particles are now at $θ_{C2} = 4π/3$ with their original velocities.
• The time interval $Δt_3$ until their next meeting will be the same as $t_1$: $Δt_3 = \frac{2\pi R}{3v} = t_1$
• Let's find the location of this third meeting by tracking P1 from $θ_{C2}$. P1 moves CCW with angular speed $ω_1 = v/R$. $Δθ_3 = ω_1 \times Δt_3 = \frac{v}{R} \times \frac{2\pi R}{3v} = \frac{2\pi}{3}$
• The position of the third meeting is: $θ_3 = θ_{C2} + Δθ_3 = \frac{4\pi}{3} + \frac{2\pi}{3} = \frac{6\pi}{3} = 2\pi$
• An angular position of 2π is equivalent to 0, which is the starting point A.

Conclusion

The particles meet again at point A after a total time of $t = t_1 + Δt_2 + Δt_3 = 3t_1$. Before they meet at point A, they have undergone two collisions:

1. The first collision at θ = 2π/3.
2. The second collision at θ = 4π/3.

The question asks for the number of elastic collisions, other than that at A, after which the particles will again reach point A. Based on our analysis, there are 2 such collisions.

Therefore, the correct answer is 2.