Method 1: Using Conservation Laws in the Lab Frame

1. Define the system and variables: Let the projectile particle have mass $M = 2m$ and initial velocity $\vec{v}_0 = v_0 \hat{i}$. Let the target particle have mass $m$ and is initially at rest, $\vec{v}_{target} = 0$. After the collision, the velocity of the projectile is $\vec{v}_1'$ at an angle $\theta$ to the initial direction. The velocity of the target is $\vec{v}_2'$ at an angle $\phi$ to the initial direction.

2. Apply Conservation of Momentum: The total momentum is conserved. We can write the conservation equations for the components along the x (initial direction) and y axes. $M v_0 = M v_1' \cos\theta + m v_2' \cos\phi$ $0 = M v_1' \sin\theta - m v_2' \sin\phi$ Substituting $M = 2m$: $2m v_0 = 2m v_1' \cos\theta + m v_2' \cos\phi \implies 2(v_0 - v_1' \cos\theta) = v_2' \cos\phi \quad (1)$ $0 = 2m v_1' \sin\theta - m v_2' \sin\phi \implies 2 v_1' \sin\theta = v_2' \sin\phi \quad (2)$

3. Apply Conservation of Kinetic Energy: The collision is perfectly elastic, so kinetic energy is conserved. $\frac{1}{2} M v_0^2 = \frac{1}{2} M (v_1')^2 + \frac{1}{2} m (v_2')^2$ Substituting $M = 2m$: $\frac{1}{2} (2m) v_0^2 = \frac{1}{2} (2m) (v_1')^2 + \frac{1}{2} m (v_2')^2$ $2 v_0^2 = 2 (v_1')^2 + (v_2')^2 \quad (3)$

4. Solve for the scattering angle $\theta$: We can eliminate $v_2'$ and $\phi$ from the equations. Square and add equations (1) and (2): $(v_2')^2 (\cos^2\phi + \sin^2\phi) = [2(v_0 - v_1' \cos\theta)]^2 + [2 v_1' \sin\theta]^2$ $(v_2')^2 = 4(v_0^2 - 2v_0 v_1' \cos\theta + (v_1')^2 \cos^2\theta) + 4(v_1')^2 \sin^2\theta$ $(v_2')^2 = 4(v_0^2 - 2v_0 v_1' \cos\theta + (v_1')^2)$ Now substitute this expression for $(v_2')^2$ into the energy conservation equation (3): $2 v_0^2 = 2 (v_1')^2 + 4(v_0^2 - 2v_0 v_1' \cos\theta + (v_1')^2)$ $2 v_0^2 = 2(v_1')^2 + 4v_0^2 - 8v_0 v_1' \cos\theta + 4(v_1')^2$ $0 = 2v_0^2 + 6(v_1')^2 - 8v_0 v_1' \cos\theta$ Rearranging this gives a quadratic equation for $v_1'$: $3(v_1')^2 - (4v_0 \cos\theta) v_1' + v_0^2 = 0$

5. Find the condition for a physical solution: For the final speed $v_1'$ to be a real, physical value, the discriminant of this quadratic equation must be non-negative ($D \ge 0$). The discriminant is $D = b^2 - 4ac$, with $a=3$, $b=-4v_0 \cos\theta$, and $c=v_0^2$. $D = (-4v_0 \cos\theta)^2 - 4(3)(v_0^2) \ge 0$ $16 v_0^2 \cos^2\theta - 12 v_0^2 \ge 0$ Since $v_0^2 > 0$, we can divide by it: $16 \cos^2\theta \ge 12$ $\cos^2\theta \ge \frac{12}{16} = \frac{3}{4}$

6. Determine the maximum angle: For a projectile heavier than the target ($M>m$), it can only be scattered in the forward direction, meaning $0 \le \theta \le \pi/2$. Thus, $\cos\theta \ge 0$. The condition $\cos^2\theta \ge 3/4$ becomes $\cos\theta \ge \frac{\sqrt{3}}{2}$. The maximum value of the angle $\theta$ corresponds to the minimum possible value of $\cos\theta$. From the inequality, the minimum value of $\cos\theta$ is $\frac{\sqrt{3}}{2}$. $\cos(\theta_{max}) = \frac{\sqrt{3}}{2}$ Therefore, the maximum angular deviation is: $\theta_{max} = \arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \text{ radians}$

Method 2: Using Center of Mass (CM) Frame

1. Velocity of the Center of Mass ($V_{CM}$): $V_{CM} = \frac{(2m)v_0 + m(0)}{2m+m} = \frac{2mv_0}{3m} = \frac{2}{3}v_0$

2. Velocities in the CM frame (initial): Projectile ($2m$): $u_1 = v_0 - V_{CM} = v_0 - \frac{2}{3}v_0 = \frac{1}{3}v_0$ Target ($m$): $u_2 = 0 - V_{CM} = -\frac{2}{3}v_0$

3. Velocities in the CM frame (final): In an elastic collision, the speeds in the CM frame remain unchanged. The particles just change direction. Let the scattering angle in the CM frame be $\theta_{CM}$. The final velocity of the projectile in the CM frame is $\vec{u}_1'$ with magnitude $|\vec{u}_1'| = u_1 = \frac{1}{3}v_0$.

4. Transform final velocity back to the Lab frame: The final velocity in the lab frame $\vec{v}_1'$ is given by $\vec{v}_1' = \vec{u}_1' + \vec{V}_{CM}$. Components of $\vec{v}_1'$: $v'_{1x} = u'_1 \cos\theta_{CM} + V_{CM} = \frac{1}{3}v_0 \cos\theta_{CM} + \frac{2}{3}v_0$ $v'_{1y} = u'_1 \sin\theta_{CM} = \frac{1}{3}v_0 \sin\theta_{CM}$

5. Find the lab scattering angle $\theta$: The scattering angle $\theta$ in the lab frame is given by $\tan\theta = \frac{v'_{1y}}{v'_{1x}}$. $\tan\theta = \frac{\frac{1}{3}v_0 \sin\theta_{CM}}{\frac{2}{3}v_0 + \frac{1}{3}v_0 \cos\theta_{CM}} = \frac{\sin\theta_{CM}}{2 + \cos\theta_{CM}}$

6. Maximize the angle $\theta$: To find the maximum $\theta$, we can maximize $\tan\theta$ with respect to $\theta_{CM}$. Let $f(\theta_{CM}) = \tan\theta$. We find the derivative and set it to zero. $\frac{d(\tan\theta)}{d\theta_{CM}} = \frac{(\cos\theta_{CM})(2 + \cos\theta_{CM}) - (\sin\theta_{CM})(-\sin\theta_{CM})}{(2 + \cos\theta_{CM})^2} = 0$ $2\cos\theta_{CM} + \cos^2\theta_{CM} + \sin^2\theta_{CM} = 0$ $2\cos\theta_{CM} + 1 = 0 \implies \cos\theta_{CM} = -\frac{1}{2}$ This value corresponds to a maximum because the second derivative is negative. If $\cos\theta_{CM} = -1/2$, then $\sin\theta_{CM} = \sqrt{1 - (-1/2)^2} = \frac{\sqrt{3}}{2}$.

7. Calculate $\theta_{max}$: Substitute these values back into the expression for $\tan\theta$: $\tan(\theta_{max}) = \frac{\sqrt{3}/2}{2 + (-1/2)} = \frac{\sqrt{3}/2}{3/2} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$ Therefore, the maximum angle is: $\theta_{max} = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \text{ radians}$

Both methods yield the same result. The correct option is D.