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, the maximum angular deviation $\theta$ of the heavier particle, as shown in the figure, in radians is:

• A. $\pi$
• B. $\tan^{-1}(\frac{1}{2})$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{6}$

Answer 1

Answer: (D)

$\frac{\pi}{6}$

Answer 2

The problem involves an elastic collision between two particles. To find the maximum angular deviation $\theta$ of the heavier particle, we can use the formula derived from conservation of momentum and kinetic energy.

Method 1: Using the center of mass frame

1. Velocity of the Center of Mass ($V_{CM}$): $V_{CM} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} = \frac{2m u_1}{2m + m} = \frac{2}{3} u_1$

2. Velocities in the Center of Mass (CM) Frame: $u_1' = u_1 - V_{CM} = u_1 - \frac{2}{3}u_1 = \frac{1}{3}u_1$ $u_2' = u_2 - V_{CM} = 0 - \frac{2}{3}u_1 = -\frac{2}{3}u_1$

3. Relating Lab Frame and CM Frame Velocities: $\vec{v_1} = \vec{v_1^{*'}} + \vec{V_{CM}}$ $v_{1x} = V_{CM} + v_1^{*'} \cos\phi = \frac{2}{3}u_1 + \frac{1}{3}u_1 \cos\phi$ $v_{1y} = v_1^{*'} \sin\phi = \frac{1}{3}u_1 \sin\phi$

4. Scattering Angle of Incident Particle ($\theta_1$): $\tan\theta_1 = \frac{v_{1y}}{v_{1x}} = \frac{\frac{1}{3}u_1 \sin\phi}{\frac{2}{3}u_1 + \frac{1}{3}u_1 \cos\phi} = \frac{\sin\phi}{2 + \cos\phi}$

5. Maximizing $\theta_1$: $\frac{d(\tan\theta_1)}{d\phi} = \frac{2\cos\phi + 1}{(2 + \cos\phi)^2} = 0 \implies \cos\phi = -\frac{1}{2}$ $\sin\phi = \pm\sqrt{1 - (-\frac{1}{2})^2} = \pm\frac{\sqrt{3}}{2}$. $\tan\theta_1 = \frac{\frac{\sqrt{3}}{2}}{2 + (-\frac{1}{2})} = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$ $\theta_1 = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$ radians.

Method 2: Using the general formula for maximum scattering angle

$\theta_{max} = \sin^{-1}\left(\frac{m_2}{m_1}\right) = \sin^{-1}\left(\frac{m}{2m}\right) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$ radians.

Therefore, the maximum angular deviation $\theta$ of the heavier particle is $\frac{\pi}{6}$ radians.