Question

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, find the maximum angular deviation θ of the heavier particle (mass 2m) in the laboratory frame, in radians.

Solution

Key Result (Standard Kinematic Result)

For a perfectly elastic collision in which a heavier particle of mass M₁ = 2m collides with a lighter stationary particle of mass M₂ = m, the maximum scattering angle of the incident (heavier) particle in the laboratory frame is given by:

$$\sin \theta_{\max} = \frac{M_2}{M_1}$$

Substituting the masses:

$$\sin \theta_{\max} = \frac{m}{2m} = \frac{1}{2}$$

Therefore:

$$\theta_{\max} = 30^\circ = \frac{\pi}{6} \text{ radians}$$

Physical Reason (Centre-of-Mass Frame Analysis)

• In the centre-of-mass (CM) frame, after a perfectly elastic collision, the two particles always move in opposite directions (180° apart).
• The velocity of the centre of mass is directed along the initial direction of the heavier particle.
• The heavier particle has smaller speed in the CM frame compared to the lighter one.
• The maximum laboratory scattering angle of the heavier particle occurs when its final velocity vector in the CM frame is perpendicular to the velocity of the CM itself.
• Geometric construction in the velocity diagram yields the famous result: sin θ_max = m₂/m₁

Special Cases Reminder

• If M₁ = M₂ → θ_max = 90°
• If M₁ > M₂ → θ_max < 90° (heavier particle cannot be scattered backward)
• If M₁ < M₂ → θ_max = 180° (heavier target can send lighter projectile backward)