= Solution
Choose positive x along the rocket's motion. If its instantaneous total <mass> is $\mu$ and it ejects a positive amount $d\eta=-d\mu$, the exhaust <velocity> in the original <inertial frame> is $v-u_0$ to first order. <Conservation of momentum> over that increment gives
$$
\mu v=(\mu-d\eta)(v+dv)+d\eta(v-u_0),\qquad \mu\,dv=u_0d\eta=-u_0d\mu.
$$
Integrating the <rocket equation> from $\mu=m+M$ to $\mu=m$ gives
$$
\boxed{v_f=u_0\ln\frac{m+M}{m}.}
$$
The constant exhaust <speed> is relative to the rocket, not the initial frame. <Kinetic energy> of rocket plus exhaust is not conserved because stored fuel <energy> is released. The calculation is Newtonian and requires resulting <speeds> small compared with the <speed of light>.
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