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 \(\theta\).
Show Hint
- \(\tan^{-1}\left(\frac{1}{2}\right)\)
- \(\frac{\pi}{6}\)
- \(\frac{\pi}{3}\)
- \(\pi\)
The Correct Option is A
Solution and Explanation
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 \[ \boxed{\theta_{\max} = \tan^{-1} \left( \frac{1}{2} \right)} \]
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