= Solution
Work in one fixed <inertial frame>, and take positive <velocity> in the rocket's direction. During $d\tau$, the rocket ejects the positive amount of <mass> $-m'(\tau)d\tau$. Its exhaust has inertial velocity $v(\tau)-u$, by the <Galilean transformation> of velocities. Once expelled, that material retains its velocity because it experiences no external <force>. The system's <momentum> is therefore the rocket momentum plus the sum of all these exhaust momenta:
$$
P(t)=m(t)v(t)-\int_0^t[v(\tau)-u]m'(\tau)\,d\tau.
$$
The minus sign comes from the positive expelled mass being $-m'(\tau)d\tau$.
Differentiate, applying the product rule to the rocket <momentum> and the <fundamental theorem of calculus> to the integral:
$$
P'=m'v+mv'-(v-u)m'=mv'+um'.
$$
The <momentum conservation> law thus gives \b[the nonrelativistic <rocket equation>]
$$
\boxed{mv'+um'=0}.
$$
The same bookkeeping for <kinetic energy> gives
$$
K(t)=\frac12m(t)v(t)^2-\frac12\int_0^t[v(\tau)-u]^2m'(\tau)\,d\tau.
$$
Its derivative is
$$
K'=mvv'+\frac12\bigl[v^2-(v-u)^2\bigr]m'
=mvv'+uvm'-\frac12u^2m'
=-\frac12u^2m',
$$
where the <rocket equation> cancels the first two terms. Thus \b[kinetic energy increases during mass ejection], with $\boxed{K'=-u^2m'/2>0}$ when $u>0$ and $m'<0$. This does not violate <conservation of energy>: the increase comes from stored <internal energy>.
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