JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Conservation of Momentum

The fragments have masses in the ratio $2:2:3$, that is $4\text{ kg}$, $4\text{ kg}$ and $6\text{ kg}$ for the $14\text{ kg}$ body. Since the body is initially at rest, the total momentum stays zero. The two equal fragments fly off perpendicular to each other at $18\text{ m/s}$, each carrying momentum $4\times 18=72\mathrm{kg}m/s$; their resultant has magnitude $\sqrt{72^2+72^2}=72\sqrt{2}\mathrm{kg}m/s$. The heavier fragment must carry an equal and opposite momentum, so $6v=72\sqrt{2}$, giving $v=12\sqrt{2}m/s$.