$A$ rocket accelerates straight up by ejecting gas downwards. In a small time interval $\Delta t$,it ejects a gas of mass $\Delta m$ at a relative speed $u$. Calculate the kinetic energy $(KE)$ of the entire system at time $t + \Delta t$ and $t$,and show that the device that ejects gas does work equal to $(\frac{1}{2}) \Delta m u^2$ in this time interval (ignoring gravity).

(A) Let $M$ be the mass of the rocket at time $t$ and $v$ be its velocity.
In time interval $\Delta t$,the rocket ejects gas of mass $\Delta m$ with a relative speed $u$.
The velocity of the ejected gas with respect to the ground is $(v - u)$.
The velocity of the rocket after time $\Delta t$ becomes $(v + \Delta v)$.
The kinetic energy of the system at time $t$ is $(KE)_t = \frac{1}{2} M v^2$.
At time $t + \Delta t$,the kinetic energy of the system is $(KE)_{t+\Delta t} = \frac{1}{2}(M - \Delta m)(v + \Delta v)^2 + \frac{1}{2} \Delta m(v - u)^2$.
Expanding this,we get $(KE)_{t+\Delta t} \approx \frac{1}{2} M v^2 + M v \Delta v - \Delta m v u + \frac{1}{2} \Delta m u^2$.
From the conservation of momentum,the thrust force $F = M \frac{dv}{dt} = u \frac{dm}{dt}$,so $M \Delta v = \Delta m u$.
Substituting $M \Delta v = \Delta m u$ into the expression for the change in kinetic energy $\Delta K = (KE)_{t+\Delta t} - (KE)_t$:
$\Delta K = (M \Delta v - \Delta m u) v + \frac{1}{2} \Delta m u^2$.
Since $M \Delta v = \Delta m u$,the first term becomes zero.
Thus,$\Delta K = \frac{1}{2} \Delta m u^2$.
By the work-energy theorem,the work done by the internal mechanism is equal to the change in kinetic energy,so $\Delta W = \frac{1}{2} \Delta m u^2$.