Three bodies A, B and C have equal kinetic energies and their masses are $$400 \text{ g}$$, $$1.2 \text{ kg}$$ and $$1.6 \text{ kg}$$ respectively. The ratio of their linear momenta is :

Three bodies A, B, C with equal kinetic energies and masses 400 g, 1.2 kg, 1.6 kg. Find the ratio of their momenta.

Recall the relation between momentum and kinetic energy. $$KE = \frac{p^2}{2m}$$, so $$p = \sqrt{2mKE}$$.

Since KE is the same for all three: $$p_A : p_B : p_C = \sqrt{m_A} : \sqrt{m_B} : \sqrt{m_C}$$

Substitute the masses. $$m_A = 0.4$$ kg, $$m_B = 1.2$$ kg, $$m_C = 1.6$$ kg. So $$= \sqrt{0.4} : \sqrt{1.2} : \sqrt{1.6}$$

Multiply all by $$\sqrt{10}$$ (equivalently, consider $$4:12:16$$ inside the roots): $$= \sqrt{4} : \sqrt{12} : \sqrt{16} = 2 : 2\sqrt{3} : 4 = 1 : \sqrt{3} : 2$$

The correct answer is Option (2): $$1 : \sqrt{3} : 2$$.