Maximum Angular Deviation in Elastic Collision of Unequal Masses

To find the maximum angular deviation $\theta$ of the incident particle in the laboratory frame, we can analyze the collision using the Center of Mass (CM) frame.

1. Velocity in the Center of Mass (CM) Frame:

Let the incident particle of mass $m_1 = 2m$ move with an initial velocity $u$ along the positive $x$-axis, and the target particle of mass $m_2 = m$ be at rest.

The velocity of the center of mass is given by: $v_{\text{cm}} = \frac{m_1 u + m_2(0)}{m_1 + m_2} = \frac{2m \cdot u}{2m + m} = \frac{2}{3}u$

In the CM frame, the initial velocity of the heavier particle is: $u_1' = u - v_{\text{cm}} = u - \frac{2}{3}u = \frac{1}{3}u$

2. Post-Collision Velocity:

Since the collision is perfectly elastic, the magnitude of the velocity of the particle in the CM frame remains unchanged after scattering: $v_1' = u_1' = \frac{1}{3}u$

Let $\Theta$ be the scattering angle of the heavier particle in the CM frame. The velocity of the particle in the laboratory frame $\vec{v}_1$ is the vector sum of the CM velocity and the velocity relative to the CM: $\vec{v}_1 = \vec{v}_{\text{cm}} + \vec{v}_1'$

The components of the laboratory velocity are: $v_{1x} = v_{\text{cm}} + v_1' \cos\Theta = \frac{2}{3}u + \frac{1}{3}u \cos\Theta$ $v_{1y} = v_1' \sin\Theta = \frac{1}{3}u \sin\Theta$

3. Maximum Angular Deviation:

Geometrically, the vector $\vec{v}_1$ is obtained by adding a vector $\vec{v}_1'$ of fixed magnitude $\frac{1}{3}u$ (which can point in any direction $\Theta$) to a fixed vector $\vec{v}_{\text{cm}}$ of magnitude $\frac{2}{3}u$.

Thus, the tip of $\vec{v}_1$ lies on a circle of radius $r = v_1' = \frac{1}{3}u$ centered at a distance $v_{\text{cm}} = \frac{2}{3}u$ from the origin along the $x$-axis.

Since $v_{\text{cm}} > v_1'$, the origin lies strictly outside this circle. The maximum angular deviation $\theta_{\max}$ in the laboratory frame occurs when the line representing the velocity vector $\vec{v}_1$ is tangent to this circle.

At the point of tangency: $\sin\theta_{\max} = \frac{v_1'}{v_{\text{cm}}} = \frac{\frac{1}{3}u}{\frac{2}{3}u} = \frac{1}{2}$

Therefore: $\theta_{\max} = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}\text{ radians}$

Hence, the correct option is (D).