Question

A mass \(2m\) collides elastically with a stationary mass \(m\). After collision, the angle \(\theta\) is formed between their velocities. Find the maximum possible value of \(\theta\).

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

Hint:

In elastic collisions where one body is initially at rest, use \(\tan \theta = \frac{m}{M}\) to find angular spread in lab frame.

Answer 1

The Correct Option is A

Assume: - Mass of incoming particle: \(M = 2m\) - Target mass at rest: \(m\) - Elastic collision: momentum and kinetic energy conserved Use result from elastic collision (scattering in lab frame): If a moving particle of mass \(M\) collides elastically with a stationary particle of mass \(m\), and both particles move after collision, then the angle \(\theta\) between their velocity vectors satisfies: \[ \tan \theta_{\text{max}} = \frac{m}{M} \Rightarrow \tan \theta_{\text{max}} = \frac{m}{2m} = \frac{1}{2} \Rightarrow \theta_{\text{max}} = \tan^{-1} \left( \frac{1}{2} \right) \]  Final Answer \[ \boxed{\theta_{\max} = \tan^{-1} \left( \frac{1}{2} \right)} \]