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. \( \tan^{-1} \left(\frac{1}{2}\right) \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{\pi}{6} \)

Hint:

For perfectly elastic collisions, use conservation of momentum in components and conservation of kinetic energy to find scattering angles.

Answer 1

The Correct Option is D

Step 1: Understand the System - A particle of mass \(2m\) collides elastically with a particle of mass \(m\) initially at rest. - We seek the maximum angular deviation \( \theta \) of the heavier particle (mass \(2m\)) after the collision. 
Step 2: Apply Conservation Laws - We use conservation of momentum (x and y) and conservation of kinetic energy (elastic collision). 
Momentum Conservation: x-direction: \[ 2mv = 2mv_1\cos\theta + mv_2\cos\phi \implies 2v = 2v_1\cos\theta + v_2\cos\phi \tag{1} \] y-direction: \[ 0 = 2mv_1\sin\theta + mv_2\sin\phi \implies 2v_1\sin\theta = -v_2\sin\phi \tag{2} \] Kinetic Energy Conservation: \[ \frac{1}{2}(2m)v^2 = \frac{1}{2}(2m)v_1^2 + \frac{1}{2}mv_2^2 \implies v^2 = v_1^2 + \frac{1}{2}v_2^2 \tag{3} \] 
Step 3: Relate Variables and Find Maximum Angle - The goal is to express \(\theta\) in terms of known constants and maximize it. This is best done with \(v_1, v_2 \) as independent variables and \(\phi \) related to \(\theta \). - Solve Eqn. (2) for \( \sin \phi \): \[ \sin \phi = -2\frac{v_1}{v_2} \sin \theta \] - Solve Eqn. (1) for \( \cos \phi \): \[ \cos \phi = \frac{2v - 2v_1 \cos \theta}{v_2} \] - Use \( \sin^2 \phi + \cos^2 \phi = 1 \): \[ \left( -2 \frac{v_1}{v_2} \sin \theta \right)^2 + \left( \frac{2v - 2v_1 \cos \theta}{v_2} \right)^2 = 1 \] \[ 4 v_1^2 \sin^2 \theta + 4(v - v_1 \cos \theta)^2 = v_2^2 \] - Use Eqn. (3) to eliminate \(v_2^2\): \[ 4 v_1^2 \sin^2 \theta + 4(v^2 - 2vv_1 \cos \theta + v_1^2 \cos^2 \theta) = 2(v^2 - v_1^2) \] \[ 4v_1^2(\sin^2\theta + \cos^2\theta) + 4v^2 - 8vv_1\cos\theta = 2v^2 - 2v_1^2 \] \[ 6v_1^2 - 8vv_1\cos\theta + 2v^2 = 0 \] \[ 3v_1^2 - 4vv_1\cos\theta + v^2 = 0 \] - Solve the quadratic for \( v_1 \) using the quadratic formula: \[ v_1 = \frac{4v\cos\theta \pm \sqrt{16v^2\cos^2\theta - 12v^2}}{6} = \frac{2v\cos\theta \pm v\sqrt{4\cos^2\theta - 3}}{3} \] - For \( v_1 \) to be real, we require: \[ 4\cos^2 \theta - 3 \geq 0 \implies \cos^2 \theta \geq \frac{3}{4} \implies |\cos\theta| \geq \frac{\sqrt{3}}{2} \] - Therefore: \[ -\frac{\pi}{6} \leq \theta \leq \frac{\pi}{6} \] - So the maximum scattering angle is \( \theta_{max} = \frac{\pi}{6} \).
Alternate Solution using the Approximation: In an elastic collision, the maximum scattering angle \( \theta_{max} \) of the heavier particle (mass \( m_1 \)) when colliding with a lighter particle (mass \( m_2 \)) is given by: \[ \sin \theta_{max} = \frac{m_2}{m_1} \] This formula is valid when \( m_1 \geq m_2 \). In this case, \( m_1 = 2m \) and \( m_2 = m \). Thus: \[ \sin \theta_{max} = \frac{m}{2m} = \frac{1}{2} \implies \theta_{max} = \frac{\pi}{6} \] Step 4: Conclusion The maximum angular deviation \( \theta \) of the heavier particle is \( \frac{\pi}{6} \). Final Answer: The correct option is:\( \boxed{\text{(D)}} \)