JEE Advanced 2025 — Paper 1 — Question 2

$2 m v_{1}=2 m v_{1 \mathrm{f}} \cos \theta+2 m v_{2 \mathrm{f}} \cos \phi$

$2 \mathrm{~m}_{1 \mathrm{f}} \sin \theta=\mathrm{mv}_{2 \mathrm{f}} \sin \phi$

$\frac{1}{2}(2 m) v_{1}^{2}+\frac{1}{2} m(0)^{2}=\frac{1}{2}(2 m) v_{1 f}^{2}+\frac{1}{2} m v_{2 f}^{2}$

$2 \mathrm{v}_{1}^{2}=2 \mathrm{v}_{1 \mathrm{f}}^{2}+\mathrm{v}_{2 \mathrm{f}}^{2}$

From (i), (ii), (iii), $3 v_{1 f}^{2}-4 v_{1} v_{1 f} \cos \theta+v_{1}^{2}=0$ $\left(-4 \mathrm{v}_{1} \cos \theta\right)^{2}-4(3)\left(\mathrm{v}_{1}^{2}\right) \geq 0$

$\cos ^{2} \theta \geq \frac{3}{4}$

$\cos ^{2} \theta \geq \frac{\sqrt{3}}{2}$

$\therefore \theta=\frac{\pi}{6}$