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JEE Advanced · Physics · 7. COM & Collisions

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:

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

Answer & Solution

Correct Answer

(D) \(\frac{\pi}{6}\)

Step-by-step Solution

Detailed explanation

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