Two particles having masses 4 g and 16 g respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is $$n : 2$$. The value of $$n$$ will be ______.

We are given two particles with masses $$m_1 = 4$$ g and $$m_2 = 16$$ g moving with equal kinetic energies. We need to find the ratio of their linear momenta.

The kinetic energy of a particle is related to its momentum by $$K = \frac{p^2}{2m}$$, which gives $$p = \sqrt{2mK}$$.

Since both particles have equal kinetic energies ($$K_1 = K_2 = K$$):

$$p_1 = \sqrt{2m_1 K} = \sqrt{2 \times 4 \times K} = \sqrt{8K}$$

$$p_2 = \sqrt{2m_2 K} = \sqrt{2 \times 16 \times K} = \sqrt{32K}$$

The ratio of their momenta is: $$\frac{p_1}{p_2} = \frac{\sqrt{8K}}{\sqrt{32K}} = \sqrt{\frac{8}{32}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$.

So $$p_1 : p_2 = 1 : 2$$. Comparing with the given ratio $$n : 2$$, we get $$n = 1$$.