Q34 / 48 · JEE Advanced 2025 (Paper 1) · Physics

In a scattering experiment, a particle of mass $2 m$ 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:
Figure

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

Solution

$2 m v_1=2 m v_{1 \mathrm{f}} \cos \theta+2 m v_{2 \mathrm{f}} \cos \phi$ $\qquad$....(i) $2 m m_{1 \mathrm{f}} \sin \theta=m v_{2 \mathrm{f}} \sin \phi$ $\qquad$....(ii) $\frac{1}{2}(2 m) v_1^2+\frac{1}{2} m(0)^2=\frac{1}{2}(2 m) v_{1 f}^2+\frac{1}{2} m v_{2 f}^2$ $2 v_1^2=2 v_{1 f}^2+v_{2 f}^2$ $\qquad$....(iii) From (i), (ii), (iii), $\begin{aligned} & 3 v_{1 \mathrm{f}}^2-4 v_1 v_{1 \mathrm{f}} \cos \theta+v_1^2=0 \\ & \left(-4 v_1 \cos \theta\right)^2-4(3)\left(v_1^2\right) \geq 0 \\ & \cos ^2 \theta \geq \frac{3}{4} \\ & \cos ^2 \theta \geq \frac{\sqrt{3}}{2} \\ & \therefore \theta=\frac{\pi}{6}\end{aligned}$

Hence, the correct answer is option: 'D'.

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