JEE Advanced 2025 Paper 1 Q02 Physics Mechanics Work-Energy-Power Hard

JEE Advanced 2025 Paper 1 · Q02 · Work-Energy-Power

In a scattering experiment, a particle of mass $2m$ collides elastically with another particle of mass $m$ that is initially at rest. Find the maximum angular deviation $\theta$ of the heavier particle (in radians).

  1. A. $\pi/2$
  2. B. $\tan^{-1}(1/2)$
  3. C. $\pi/3$
  4. D. $\pi/6$
Reveal answer + step-by-step solution

Correct answer:D

Solution

For elastic scattering of mass $m_1$ on stationary mass $m_2$, when $m_1 > m_2$ the maximum lab-frame scattering angle of the projectile is given by $\sin(\theta_{\max}) = m_2/m_1$. Here $m_1 = 2m$, $m_2 = m$, so $\sin(\theta_{\max}) = 1/2 \implies \theta_{\max} = \pi/6$. (Standard derivation: in the CM frame the projectile scatters by any angle $\phi$; the lab angle satisfies $\tan(\theta) = \sin(\phi)/(\cos(\phi) + m_1/m_2)$. Maximize over $\phi$ to get $\sin(\theta_{\max}) = m_2/m_1$.) Answer: $\pi/6$.

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