In a scattering experiment, a particle of mass $2m$ collides with another particle of mass $m$, which is initially at rest. Assuming the collision to be perfectly elastic, the maximum angular deviation $\theta$ of the heavier particle, as shown in the figure, in radians is:

Let the heavier particle (mass 2m) come in with speed u along +x. The centre of mass moves at V_cm = 2mu/3m = 2u/3 along +x.

In the CM frame the heavy particle moves at v₁′ = u − 2u/3 = u/3 and the light one at v₂′ = 2u/3 in the opposite direction; their momenta are equal and opposite. In an elastic collision with zero total momentum both speeds are unchanged — the collision only rotates the line of motion by an angle Θ, which can take any value.

Back in the lab, v₁ = V_cm + v₁′: a fixed vector of magnitude 2u/3 along +x plus a vector of magnitude u/3 in any direction. The tip of v₁ therefore lies on a circle of radius u/3 centred at distance 2u/3 from the origin. The deflection θ is greatest when v₁ is tangent to this circle:

sin θ_max = (u/3)/(2u/3) = 1/2, so θ_max = π/6 (at Θ = 120°).

Since u/3 < 2u/3 the circle does not reach the origin, so the heavier particle can never be turned back: π is impossible. In general sin θ_max = m₂/m₁ for m₁ > m₂.

Answer: θ_max = π/6 — option (4).