Two particles of equal mass m start moving simultaneously from the same point on a horizontal circular track of radius R . They move in opposite directions with the same constant speed v . What is the angular distance covered by each particle from the starting point at the instant when the magnitude of the total linear momentum of the two-particle system exactly equals the momentum of a single particle?

Step-by-step solution

Let the particles start from the position (0, R) on the circular track. At any instant, let each particle have covered an angular distance . The velocity of the first particle (moving clockwise) is v ₁ = v i - v j . The velocity of the second particle (moving counter-clockwise) is v ₂ = -v i - v j . The total linear momentum of the system is P = m v ₁ + m v ₂ = -2mv j . The magnitude of the total momentum is P = 2mv . We are given that P equals the momentum of a single particle, which is mv . Therefore, 2mv = mv =