The relationship between kinetic energy ($K$) and linear momentum ($p$) for a body of mass $m$ is given by $K = \frac{p^2}{2m}$, which implies $p = \sqrt{2mK}$. Let the initial kinetic energy be $K_1$. The initial momentum is $p_1 = \sqrt{2mK_1}$. If the kinetic energy is increased by $300\%$, the final kinetic energy $K_2$ becomes: $K_2 = K_1 + \frac{300}{100}K_1 = K_1 + 3K_1 = 4K_1$ The final momentum $p_2$ corresponding to this new kinetic energy is: $p_2 = \sqrt{2m(4K_1)} = 2\sqrt{2mK_1} = 2p_1$ The percentage increase in momentum is: $\text{Percentage increase} = \frac{p_2 - p_1}{p_1} \times 100\% = \frac{2p_1 - p_1}{p_1} \times 100\% = 100\%$.