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:

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