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(B) Let the initial mass be $m$ and initial velocity be $v$. The initial kinetic energy is $(K.E.)_i = \frac{1}{2}mv^2$ and initial momentum is $p_i =

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$2 \sqrt{2}$

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(B) Let the initial mass be $m$ and initial velocity be $v$. The initial kinetic energy is $(K.E.)_i = \frac{1}{2}mv^2$ and initial momentum is $p_i = mv$.After the fuel is burned, the final mass is $m' = \frac{m}{2}$ and the final kinetic energy is $(K.E.)_f = 16 	imes (K.E.)_i = 16 	imes \frac{1}{2}mv^2 = 8mv^2$.Let the final velocity be $v'$. Then $(K.E.)_f = \frac{1}{2}m'v'^2 = \frac{1}{2}(\frac{m}{2})v'^2 = \frac{1}{4}mv'^2$.Equating the two expressions for $(K.E.)_f$: $\frac{1}{4}mv'^2 = 8mv^2 \Rightarrow v'^2 = 32v^2 \Rightarrow v' = \sqrt{32}v = 4\sqrt{2}v$.The final momentum is $p_f = m'v' = (\frac{m}{2})(4\sqrt{2}v) = 2\sqrt{2}mv$.The factor by which momentum increases is $\frac{p_f}{p_i} = \frac{2\sqrt{2}mv}{mv} = 2\sqrt{2}$.

Consider a rocket is being fired. The kinetic energy of the rocket is increased by $16$ times, whereas its total mass is reduced by half through the burning of fuel. The factor by which its momentum increases is

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