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(A) The relationship between linear momentum $p$ and kinetic energy $E$ is given by $p = \sqrt{2mE}$.Since mass $m$ is constant,$p \propto \sqrt{E}$.L

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$100$

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(A) The relationship between linear momentum $p$ and kinetic energy $E$ is given by $p = \sqrt{2mE}$.Since mass $m$ is constant,$p \propto \sqrt{E}$.Let the initial kinetic energy be $E_1 = E$.The final kinetic energy is $E_2 = E + 300\% 	ext{ of } E = E + 3E = 4E$.The initial momentum is $p_1 = \sqrt{2mE}$.The final momentum is $p_2 = \sqrt{2m(4E)} = 2\sqrt{2mE} = 2p_1$.The percentage increase in momentum is given by $\frac{p_2 - p_1}{p_1} 	imes 100\%$.Substituting the values: $\frac{2p_1 - p_1}{p_1} 	imes 100\% = \frac{p_1}{p_1} 	imes 100\% = 100\%$.

If the kinetic energy of an object is increased by $300 \%$,find the percentage $(\%)$ increase in its linear momentum.

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