Question 2

Maximum scattering angle of a heavier projectile (mass M) off a lighter target (mass m) at rest in an elastic collision: sin θ_max = m/M. Here M = 2m, so sin θ_max = 1/2 → θ_max = π/6. **Derivation sketch:** In the CM frame, the heavy particle's speed is v_cm-relative = m·u/(2m+m) = u/3, and the CM velocity is 2mu/(3m) = 2u/3. The lab velocity of the heavy particle is the vector sum of the CM velocity (2u/3) and a vector of magnitude u/3 in any direction; the maximum deflection occurs when this is tangent, giving sin θ_max = (u/3)/(2u/3) = 1/2, so θ_max = π/6.

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